ROUTERA


Capacitors

Class 12th Concepts Of Physics Part 2 HC Verma Solution



Short Answer
Question 1.

Suppose a charge + Q1 is given to the positive plate and a charge –Q2 to the negative plate of a capacitor. What is the “charge on the capacitor”?


Answer:

Given:


Charge on positive plate=Q1


Charge on negative plate=Q2



Assume a rectangular gaussian surface ABCD as shown in fig.


Let the charge on the capacitor plates be “q” and the area of plates be A. Then,


Charge appearing on face 1=Q1-q.


Charge appearing on face 2=q.


Similarly,


Charge appearing on face 3= -q.


Since, the total charge enclosed by a closed surface =0)


Charge appearing on face 4=Q2 +q.


Formula used:


We know that,


I) Electric field inside any conductor=0.


∴ Electric field at point Pinside plate)=0.


E1+E2+E3+E4=0 …i)


This Electric field is the net effect of fields at point P due to faces I, II, III and IV.


II) Electric field due a thin sheet, E=


Where


E is the electric filed due to thin plate


Q is the total charge enclosed in the gaussian surface


A is the area of the plate


εo is the permittivity of the vacuum


Thus, Electric field at point P due to face I E1=


Electric field at point P due to face II E2=


Electric field at point P due to face III E3=


Electric field at point P due to face IV E4=


negative sign because electric field due to face IV is in leftwards direction).


Putting the values in equation (i) we get,



On solving the above equation, we get


Q1-q+q-q-Q2-q=0


Q1-Q2-2q=0


q


Thus, the charge on the capacitor is



Question 2.

As can you say that the capacitance C is proportional to the charge Q?


Answer:

We know that,


Potential difference V is the work done per unit positive charge in taking a small test charge from conductor 2 to 1 against the field. Consequently, V is also proportional to Q and the ratio Q/V is a constant C known as capacitance of the capacitor.




The value of this capacitance depends only on the size, shape and position of conductor and its plates and not on the potential difference applied by the battery or th charge on the plates.


For example: the capacitance in case of an isolated spherical capacitor is given by



Where,


R=radius of the spherical conductor.


∴ Capacitance cannot be said to be dependent on charge Q.


Thus, capacitance of the capacitor is independent of the charge on the capacitor.



Question 3.

A hollow metal sphere and a solid metal sphere of equal radii are given equal charges. Which of the two will have higher potential?


Answer:

We know that,


Charge given to any conductor appearsentirely on its outer surfaceevenly.


Therefore, if equal amount of charge Q are given to a hollow and solid spheres, the entire charge Q will appear on their spherical surfaces and since they both have equal radius, capacitance of both spheres are given by


C=4πϵ0 R


Where


R= radius of the spherical capacitor.


Now, using


We get


∴ Potential of both the spheres hollow and solid) will be same.



Question 4.

The plates of a parallel-plate capacitor are given equal positive charges. What will be the potential difference between the plates? What will be the charges on the facing surfaces and on the outer surfaces?


Answer:

Given:


Two plates of a parallel plate capacitor with equal charge.


Here, both the plates are given same charge +Q.



Consider q charge on face II so that induced charge on face III is -q


Assume a rectangular Gaussian surface ABCD having area, A as shown in the above fig.


Thus, Electric field at point P due to face I E1=


Electric field at point P due to face II E2=


Electric field at point P due to face III E3=


Electric field at point P due to face IV E4=


negative sign because electric field due to face IV is in leftwards direction).


Since , point P lies inside the conductor thee total electric field at P must be zero


∴ E1+E2+E3+E4=0


=0


∴ q=0


Therefore zero charge appears on face II and III and Q charge appears on face I and IV


Potential difference b/w the plates is given by



∴ V=0 both the plates are at same potential since both are given equal charges)


1)Potential difference between the plates=0.


2)Charges on outer faces of plates=+Q.


3)Charges on inner faces of plates=0.



Question 5.

A capacitor has capacitance C. Is this information sufficient to know what maximum charge the capacitor can contain? If yes, what is this charge? If no, what other information is needed?


Answer:

Charge on the capacitor is given by product of capacitance and potential difference across capacitor plates


Charge on the capacitor,


Where,


C is the capacitance of the capacitor


V is the potential on the capacitor


Therefore, without knowing the potential difference and only capacitance we cannot find out the maximum charge capacitor can contain.


∴ The following information is insufficient.



Question 6.

The dielectric constant decreases if the temperature is increased. Explain this in terms of polarization of the material.


Answer:

When a polar or non polar material is placed in an external electric field , the electron charge distribution inside the material is slightly shifted opposite to the electric field and this induces a dipole moment in any volume of the material.


The polarization vector Pis defined as this dipole moment per unit volume.


When a dielectric rectangular slab is placed in an external electric field the dipoles get aligned along the field and the right and left surfaces of slab gets positive and negative charges as shown in fig. known as induced charge.



The more the dipoles are aligned with the external field , the more the dipole moment and thus more is the polarization.


Because of these induced charges an extra electric field is produced inside the material opposite to the direction of external field and the net electric field is given by



Where ,


K is the constant for a given dielectric known as dielectric constant of the dielectric >1)


E0 is the field in vacuum.


On increasing temperature, the random motion of molecules or dipoles increases due to thermal agitation and the dipoles get less aligned with the electric field and thus dipole moment decreases.


Since polarization is given by dipole moment per unit volume, it also decreases


And since ,dielectric constant is described by the polarization of the material


Thus, on increasing temperature, dielectric constant decreases.



Question 7.

When a dielectric slab is gradually inserted between the plates of an isolated parallel-plate capacitor, the energy of the system decreases. What can you conclude about the force on the slab exerted by the electric field?


Answer:

When a dielectric slab is gradually inserted between the plates of an isolated parallel-plate capacitor, the energy of the system come out to be a linear function of xdisplacement of the slab inside capacitor measured from the center of the plate).


since x decreases, the energy of the system decreases.


We know that



Where


is the rate of change of potential energy function with x


-ve sign indicates that force is in negative direction when energy increases with respect to x)


Therefore, Force on the slab exerted by the electric field is constant and positive.




Objective I
Question 1.

A capacitor of capacitance C is charged to a potential V. The flux of the electric field through a closed surface enclosing the capacitor is
A.

B.

C.

D. zero


Answer:

Given: a capacitor of capacitance C charged to a potential V

Gauss’s law:


Electric flux ϕ) through a closed surface S is given by



Where,


Q is the charge enclosed by S


εo is the permittivity of the free space


According to the gauss law


Electric flux,


Where


Q is the total charge enclosed in the gaussian surface


εo is the absolute permittivity of the vacuum


Since, the two plates of capacitor contains equal and opposite charges


∴ Total charge enclosed by the surface ⇒ Q-Q=0


Putting the values of total charge in gauss law, we get





∴ the electric flux through the closed surface enclosing the capacitor=0.


Question 2.

Two capacitors each having capacitance C and breakdown voltage V are joined in series. The capacitance and the breakdown voltage of the combination will be
A. 2C and 2V

B. C/2 and V/2

C. 2C and V/2

D. C/2 and 2V


Answer:

Two capacitance each having capacitance C and breakdown voltage V joined in series.


The general formula for effective capacitance of a series combination of n capacitors is given by



The equivalent capacitance of two capacitors in series is given by



Where


C1 is the capacitance of the first capacitor


C2 is the capacitance of the second capacitor


On Solving for C, we get



Since, C1=C2=C


Putting the values of C1 and C2, we get



Also, Capacitors in series have same amount of charge


∴ Q1=Q2=Q


Therefore, potential difference across both the capacitors are also equal to V


So, the voltage across the system is the sum of voltage across each capacitor.


Therefore, the breakdown voltage of the combination =V+V2V


Thus, the capacitance and breakdown voltage of the combination is C/2 and 2V respectively


Question 3.

If the capacitors in the previous question are joined in parallel, the capacitance and the breakdown voltage of the combination will be
A. 2C and 2V

B. C and 2V

C. 2C and V

D. C and V


Answer:

Given:

Two capacitors of capacitance C each and breakdown voltage V connected in parallel



Explanation:


The general formula for effective capacitance of a series combination of n capacitors is given by



The equivalent capacitance of two capacitors connected in parallel are given by



Where


C1 is the capacitance of the capacitor C1


C2 is the capacitance of the capacitor C2


Here C1=C2=C


Putting the values in the above formula, we get


Ceq=C+C=2C


Since, potential difference across capacitors in parallel are equal


Therefore voltage across the system is equal to the voltage across a single capacitor.


Therefore, breakdown voltage of the combination =V


Question 4.

The equivalent capacitance of the combination shown in figure is



A. C

B. 2C

C. C/2

D. none of these


Answer:


Since the both ends of the capacitor on the right is connected at same point.


Both the plates of the capacitor are at same potential and potential difference across capacitor becomes 0


∴ V=0


And the capacitor C on the right now becomes useless and


Thus, capacitor is replaced by a short circuit.


The circuit now becomes



Which involve two equal capacitors of capacitance C connected in parallel.



Thus, the equivalent capacitance of the two capacitor in parallel combination is


Ceq=C1+C2


Since C1=C2=C


Putting the value in the above formula, we get


Ceq=C+C=2C


Therefore, equivalent capacitance of the combination is C+C=2C.


Thus, the equivalent capacitance of the combination is 2C


Question 5.

A dielectric slab is inserted between the plates of an isolated capacitor. The force between the plates will
A. increase

B. decrease

C. remain unchanged

D. become zero


Answer:

We know that,

Force between the plates of the capacitor is given by



Where,


Q=charge on the capacitor


A=area of plates


Derivation:


Suppose charge Q and -Q are provided on plates of capacitor of area A.


Since, F=QE


Where


Q is the test charge on the point charge


E is the electric field intensity.


Field due to charge Q on one plate is



Force on the plate with charge -Q will be



Considering magnitude, each plate applies a force of


Each.


On increasing a dielectric slab between the plates of the capacitor, the charge on the plates remains constant as the plates are isolated) .


Since, area of plates does not change, force between the plates remain constant.


Question 6.

The energy density in the electric field created by a point charge falls of with the distance from the point charge as
A. 1/r

B. 1/r2

C. 1/r3

D. 1/r4


Answer:

We know that,

The energy stored per unit volumeenergy density) in an electric field E is given by



Where,


E=magnitude of electric field intensity


ϵ0=absolute permittivity of vacuum


As we know that,


And the electric field due to a point charge Q at a distance r is given by



Therefore, energy densityby formula)


And E




Thus , the energy density in the electric field created by a point charge falls of with distance from a point charge as


Question 7.

A parallel-plate capacitor has plates of unequal area. The larger plate is connected to the positive terminal of the battery and the smaller plate to its negative terminal. Let Q+ and Q be the charges appearing on the positive and negative plates respectively.
A. Q+ > Q

B. Q+ = Q

C. Q+ < Q

D. The information is not sufficient to decide the relation between Q+ and Q


Answer:

Given:

A parallel plate capacitor with plates of unequal area and charges on larger and smaller plates are Q+ and Q- respectively



Explanation:


When you have two plates of unequal areas facing each other , the electric field is present only in their common area ignoring fringe effects.)


Therefore, charges acquire only on the facing common areas of the plates of the capacitor. So, if the plates have unequal area it doesn’t matter as only the common facing area of both the plates acquire charges


Therefore, we are left with a capacitor with plates area A where A is the common area



Question 8.

A thin metal plate P is inserted between the plates of a parallel-plate capacitor of capacitance C in such a way that its edges touch the two plates figure. The capacitance now becomes.



A. C/2

B. 2C

C. 0

D. indeterminate ∞)


Answer:

Given: a parallel plate capacitor with a thin metal plate P inserted in between such that it touches the two plates.

Explanation:


When two plates of a capacitor are connected by a conductor) redistribution of charge takes place and both plates acquire same potential.


Thin metal plate P is a conductor and when connecting it to both plates of capacitor, charges gets neutralized and both the plates acquire same potential.


potential difference = 0


We know that,



Where,


V is the potential difference across capacitor


Q=charge on the capacitor



∴ Capacitance of the capacitor becomes infinite and it can hold any amount of charge.


Thus, a thin metal plate p is inserted between the plates of a parallel plate capacitor of capacitance C in such a way that its edge touch the two plates. The capacitance now becomes ∞.


Question 9.

Figure shows two capacitors connected in series and joined to a battery. The graph shows the variation in potential as one moves from left to right on the branch containing the capacitors.



A. C1 > C2

B. C1 = C2

C. C1 < C2

D. The information is not sufficient to decide the relation between C1 and C2.


Answer:

We know that,

Voltage dropor potential difference) across capacitor is given by



Where,


Q=charge on the capacitor


C=capacitance of the capacitor


By looking at the graph,



We can see that first increment in voltage is greater than the second increment. Therefore,


we can conclude that voltage drop across capacitor C1 is greater than the voltage drop across capacitor C2


on moving left to right C1 comes first)


Since charges on the capacitors in series are same ,


∴ Q1=Q2


The potential drop across the capacitor C1 is more than Capacitor C2.


V1>V2


Putting the values of V, we get



On solving, we get



Or


C2>C1


Thus, the capacitance of the capacitor C1 is less than C2


Question 10.

Two metal plates having charges Q, –Q face each other at some separation and are dipped into an oil tank. If the oil is pumped out, the electric field between the plates will
A. increase

B. decrease

C. remain the same

D. become zero


Answer:

Dielectric constant of a substance is the factor>1) by which the capacitance increases from its vacuum value, when the dielectric is inserted fully between the plates of the capacitor


Where,


C=capacitance in presence of dielectric


C0=capacitance in presence of vacuumK=1)


The electric field between the plates of a capacitor when the space between the plates is filled with a dielectric of dielectric constant K is given by



Where,


Q=charge on the capacitor


A=area of metal plates


K=dielectric constant


ϵ0=permittivity of vacuum


When dipped in oil tank value of K>1


When oil is removed there is air between the plates with K~1


∴ the value of K decreases when oil is pumped out


By the formula,



So as K decrease from greater than 1 to 1 , the electric field increases.


Therefore, after pumping out oil, the electric field between the plates increases.


Question 11.

Two metal spheres of capacitances C1 and C2 carry some charges. They are put in contact and then separated. The final charges Q1 and Q2 on them will satisfy.
A.

B.

C.

D.


Answer:

Given: two metal spheres of capacitances C1 and C2 carrying some charges.

Explanation:


Two metal spheres carrying different charges have different electric fields on their surfaces and have different potential. When they are put in contact , due to potential difference, charge transfer takes place between them such that they acquire same potential .


When two conductors are placed in contact with each other they acquire same potential.


Potential difference V across capacitor is given by the formula



Where


Q=charge on the capacitor


C=capacitance of the capacitor


Therefore, after putting them in contact and separating them, if the final charges are given by Q1 and Q2 then




Question 12.

There capacitors of capacitances 6 μF each are available. The minimum and maximum capacitances, which may be obtained are
A. 6 μF, 18 μF

B. 3 μF, 12 μF

C. 2 μF, 12 μF

D. 2 μF, 18 μF


Answer:

Given:

Three capacitors of capacitances 6μF each


i)The minimum capacitance can be obtained by connecting all three capacitors in series.


In series combination, charges on the two plates are same on each capacitor.


The general formula for effective capacitance of a series combination of n capacitors is given by



The equivalent capacitance in this case is given by





ii) The maximum capacitance can be obtained by connecting all three capacitors in parallel.


In this case, the same potential difference is applied across all capacitors.


The general formula for effective capacitance Ceq for parallel combination of n capacitors is given by



The equivalent capacitance in this case is given by



⇒ Ceq=3C


Ceq=3× 6=18μF


Therefore, the maximum and minimum capacitance that can be obtained is 18μF and 2μF respectively.



Objective Ii
Question 1.

The capacitance of a capacitor does not depend on
A. the shape of the plates

B. the size of the plates

C. the charges on the plates

D. the separation between the plates.


Answer:

We know that,

Capacitance of a capacitor only depends on shape, size and geometrical placing.


For example:


Capacitance of a parallel plate capacitor is given by



Where,


A=area of cross-section of plates


K=dielectric constant


ϵ0=absolute permittivity of medium


d=distance between the plates


∴ It does not depend on charges on the plates


So, The capacitor does depends on the shape and size of the plates and separation between the plates.


Question 2.

A dielectric slab is inserted between the plates of an isolated charged capacitor. Which of the following quantities will remain the same?
A. The electric field in the capacitor

B. The charge on the capacitor

C. The potential difference between the plates

D. The stored energy in the capacitor


Answer:

Since, the capacitor is isolated, it has no connections to any battery. Therefore, it is not possible to exchange charge due to absence of any external voltage source.

On inserting a dielectric slab of dielectric constant K, capacitance will change to KC.


Since, Charge remains constant and capacitance changes, voltage will also change according to the formula


Energy stored in the capacitor is given by



Where


Q is the charge on the capacitor


C is the capacitance of the capacitor


Derivation:


Initially consider two uncharged conductors 1 and 2 . We are transferring charge from conductor 2 to 1 such that at the end 1 gets charge Q and 2 gets charge -Q



Consider an intermediate stage where conductors 1 and 2 have charges Q’ and -Q’ respectively. At this stage potential difference V’ between conductors is given by Q’/C where C is the capacitance of the system.


For transferring a small charge dQ’ from 2 to 1 work done is given by



The total energy stored in the capacitor is summation of all these works done in transferring charge from 0 to Q.



Therefore total energy stored in capacitor is given by



Where,


Q=charge on the capacitor


C=capacitance of the capacitor


Which also changes due to change in capacitance.


Therefore on inserting dielectric slab between the plates of an isolated charge capacitor the charge on the capacitor does not change.


Question 3.

A dielectric slab is inserted between the plates of a capacitor. The charge on the capacitor is Q and the magnitude of the induced charge on each surface of the dielectric is Q’.
A. Q’ may be larger than Q.

B. Q’ must be larger than Q.

C. Q’ must be equal to Q.

D. Q’ must be smaller than Q.


Answer:

When a dielectric slab of dielectric constant K is introduced between the plates of the capacitor, the net electric field in the dielectric becomes


Where,


E0 is the electric field when there is vacuum between the plates.


K is the dielectric constant of the dielectric


The net electric field is due to charges +Q, -Q and due to induced charges +Q’,-Q’in opposite direction).


So, the net electric field becomes




Where,


Q’=induced charge due to dielectric


Q=charge on the capacitor in vacuum


K=dielectric constant


Since dielectric constant K>1


Q’<Q


Therefore, on inserting a dielectric slab between plates of capacitor the induced charge Q’ is less than Q.


Question 4.

Each plate of a parallel plate capacitor has a charge q on it. The capacitor is now connected to a battery. Now,
A. the facing surfaces of the capacitor have equal and opposite charges.

B. the two plates of the capacitor have equal and opposite charges

C. the battery supplies equal and opposite charges to the two plates

D. the outer surfaces of the plates have equal charges.


Answer:

When battery is not connected, the outer surfaces will have charge +q and inner faces of the plates will have zero charge each.


When a battery is connected to the plates of the capacitor the charges on the plate redistribute in such a way that the potential difference between the plates becomes equal to the emf of the battery.


So, the inner surfaces will have equal and opposite charges according to Q=CV


Where C is the capacitance and V is the emf of the battery. Thus we can say that the battery supplies equal and opposite charges CV) to two plates.


And the charges on the outer surfaces remain same as on connecting the battery only charges are transferred and total charge remains constant so to have zero field inside plate the outer face charges have to be same.


Therefore when a parallel plate capacitor with each plate having charge q is connected to a battery then the facing surfaces have equal and opposite charge and the outer surface will have equal charge.


Question 5.

The separation between the plates of a charged parallel-plate capacitor is increased. Which of the following quantities will change?
A. Charge on the capacitor

B. Potential difference across the capacitor

C. Energy of the capacitor

D. Energy density between the plates.


Answer:

When we increase the separation between the plates of a charged parallel capacitor the value of Capacitance decreases by the formula


Where,


d=separation between the plates


A=area of plates


ϵ0=absolute permittivity of vacuum


Charge on the capacitor remains unchanged because no charge transfer takes place.


∴ Potential difference across the capacitor changes by the formula



Where,


Q= charge on the capacitor


C=capacitance


Similarly Energy across the capacitor given by



Where,


C=capacitance


V=potential difference across capacitor


So, as V changes energy stored also changes


Therefore, on increasing separation between the plates of capacitor, potential difference and energy of capacitor changes whereas charge and energy density remains the same.


Question 6.

A parallel-plate capacitor is connected to a battery. A metal sheet of negligible thickness is placed between the plates. The sheet remains parallel to the plates of the capacitor.
A. The battery will supply more charge.

B. The capacitance will increase.

C. The potential difference between the plates will increase.

D. Equal and opposite charges will appear on the two faces of the metal plate.


Answer:

Let E0=V0/d be the electric field between the plates when there is no electric and the potential difference is V0. If the dielectric of dielectric constant K is now inserted , the electric field in the dielectric will be

The potential difference will then be



Where,


t=thickness of dielectric slab


d=separation between the plates of capacitor


Here, since metal plate is of negligible thickness , t=0


And V=E0d=V0


Thus the potential remains same c) is incorrect) and the charge Q0 on plates also remains same.


Therefore



∴ capacitance remains same.b) is incorrect)


Since charge on the capacitor remains same, no extra charge is supplied by the batterya) is incorrect)


We know that when dielectric is introduced between the plates of capacitor this polarized dielectric is equivalent to two charged surfaces with induced surface charges Q’ and -Q’


Here, the dielectric is the metal plate and therefore equal and opposite charges appear on the two faces of metal plate.


Question 7.

Following operations can be performed on a capacitor:

X – connect the capacitor to a battery of emf ϵ.

Y – disconnect the battery.

Z – reconnect the battery with polarity reversed.

W – insert a dielectric slab in the capacitor

A. In XYZ perform X, then Y, then Z) the stored electric energy remains unchanged and no thermal energy is developed.

B. The charge appearing on the capacitor is greater after the action XWY than after the action XYW.

C. The electric energy stored in the capacitor is greater after the action WXY than after the action XYW.

D. The electric field in the capacitor after the action XW is the same as that after WX.


Answer:

Option b) is correct because when a dielectric slab W is inserted in the capacitor in the presence of a battery the capacitance increases by a factor of Kdielectric constant)


Where


K is the dielectric constant


εo is the permittivity of the free space


A is the area of the plate,


d is the distance between the plates of the capacitor,


As the capacitance increases with the insertion of the dielectric, the charge appearing on the capacitor increases


Since,


Q=CV


Where,


Q is the charge on the capacitor


C is the capacitance of the capacitor


V is the potential difference supplied by the battery


Whereas capacitance does not change in case of inserting slab after removing the battery.


Optionc) is correct as


In process WXY after inserting a dielectric slab in the capacitor, the capacitance becomes



Where C0 is the capacitance in a vacuum and K is the dielectric constant. Now if the capacitor is connected to the battery of emf ϵ, then potential difference across the capacitor is given by ϵ, and the stored electrical energy is given by



Whereas in process XYW the energy is given by



Therefore, The electric energy stored in the capacitor is greater after the action WXY than after the action XYW.


Option→d) is correct because in both cases Electric field in the capacitor reduces to


Where


E0=electric field in c=vacuum


K=dielectric constant


note that it does not matter whether the battery is connected afterwards or before in 4th part)


The electric field in the capacitor after the action XW is the same as that after WX.



Exercises
Question 1.

When 1.0 × 1012 electrons are transferred from one conductor to another, a potential difference of 10V appears between the conductors. Calculate the capacitance of the two-conductor system.


Answer:

Given:


n=1.0 × 1012


V= 10 V


Formula used:


1. Charge can be given by the formula



Where Q→ Charge on the capacitor


n → number of the electrons


e → electric charge of an electron =


2. The capacitance of a parallel plate Capacitor is given by



Where Q → charge on the capacitor


V → Voltage or potential difference


Putting the values in the formula 1, we get



Substitute Q and C in Formula 2), we get



Thus, the capacitance of the parallel plate capacitor is


So, if 1.0 × 1012 electrons are transferred between two conductors the capacitance of the parallel plate capacitor is F when a potential difference is 10V



Question 2.

The plates of a parallel-plate capacitor are made of circular discs of radii 5.0 cm each. If the separation between the plates is 1.0 mm, what is the capacitance?


Answer:

Parallel plate capacitor: When two conducting plates are connected in parallel and separated by some distance then parallel plate capacitor will be formed.


Given:



Radius r= 5cm


Separation between the plates is


Formula used:


Capacitance is given by the formula



Where,


C is the capacitance of the capacitor


εₒ is the permittivity of free space =


A is the area of the circle m2


Because capacitor plates are made of circular discs)


d is the separation between the plates


Putting the values in the above relation, we get



Capacitance is of a circular disc parallel plate capacitor




Question 3.

Suppose, one wishes to construct a 1.0-f capacitor using circular discs. If the separation between the discs be kept at 1.0 mm, what would be the radius of the discs?


Answer:

Given:


C = 1 F


d=1 mm


Formula used :


Capacitance is given by



Where


C is the capacitance of the capacitor


D is the separation between the capacitor plates


A is the area of a circular plate capacitor


ε₀ is the permittivity of the free space=8.85×10-12 F/m


A = area of the circle =.Because capacitor plates are circular discs.






=6×103 m=6000 m=6 km


For the construction of 1F capacitor with 1mm separation, we need to take the radius r=6 Km.



Question 4.

A parallel-plate capacitor having plate area 25 cm2 and separation 1.00 mm is connected to a battery of 6.0V. Calculate the charge flown through the battery. How much work has been done by the battery during the process?


Answer:

Given


Area, A=25 cm2 =25×10-4 m2


Voltage, V=6V


The separation between the plates is d= 1mm=1×10-3



Formula used:


When a capacitor is connected to a capacitor, the charge can be calculated



Where


Q is the charge of the capacitor


C is the capacitance of the capacitor


V is the Voltage or potential difference across the plates of the capacitor.


Capacitance C can be calculated by the formula



Where


C is the capacitance of the capacitor


D is the separation between the capacitor plates


A is the area of a circular plate capacitor


ε₀ is the permittivity of the free space,




When the capacitor is connected to a 6V battery, Charge flow through the battery is the same as the charge that can be withstand with the capacitor.


Substitute the value of C in 1)




Work is done by the battery



Where


W is the work done by the battery


Q is the charge on the plates of the capacitor


V is the voltage across the plates of the capacitor





Charge flows through the battery is and work done by the battery is =8×10-10 J



Question 5.

A parallel-plate capacitor has plate area 25.0 cm2 and separation of 2.00 mm between the plates. The capacitor is connected to a battery of 12.0V.

a) Find the charge on the capacitor.

b) The plate separation is decreased to 1.00 mm. Find the extra charge given by the battery to the positive plate.


Answer:

a) Given Area


Voltage V=12V


Separation between the plates d=2mm


Formula used :


Charge of the capacitor can be calculated as



The capacitance of the parallel plate capacitor is given by



Where,


C is the capacitance of the parallel plate capacitor


D is the separation between the capacitor plates


A is the area of a circular plate capacitor


ε₀ is the permittivity of the free space, ε₀=8.85×10-12 F/m







b) d is decreased to 1.0mm. We have to calculate the extra charge given by the battery to the positive plate.






Extra charge is =


Charge on the capacitor when d = 2mm is =


When d is decreased to 1.00 mm the extra charge given by the battery is =



Question 6.

Find the charges on the three capacitors connected to a battery as shown in figure. Take C1 = 2.0 μF, C2 = 4.0 μF, C3 = 6.0 μF and V = 12 volts.




Answer:

Redraw the circuit given



Given


Capacitance of C1 capacitor =2μF


Capacitance of C2 Capacitor =4μf


Capacitance of C3 capacitor =6μF


Voltage V=12V


Formula used


All three capacitors are in parallel. When capacitors are in parallel, we will add them.


Equivalent Capacitance



Charge of a capacitor can be calculated by the for formula



Where


Q is the charge of the capacitor


Ceq is the equivalent Capacitance of the capacitor


V is the voltage or potential difference across the plates of the capacitor


Charge on capacitor C1 is


Coulombs


Charge on capacitor C2 is



Charge on capacitor C3 is



Charge on capacitors 2μF, 4μF and 6μF are 24C, 48C, 72C respectively.



Question 7.

Three capacitors having capacitances 20 μF, 30 μF and 40 μF are connected in series with a 12 V battery. Find the charge on each of the capacitors. How much work has been done by the battery in charging the capacitors?


Answer:

Given


Capacitance of the C1 capacitor = 20μF


Capacitance of the C2 capacitor = 30μF


Capacitance of the C3 capacitor = 40μF


All the Capacitors are connected in series.



Formula used :


Equalent capacitance is







Capacitors are connected in series, so the charge on each of them is the same .


Formula used:


Charge can be calculated as



Where


Q is the charge on the capacitor


Ceq is the equivalent Capacitance


V is the voltage




Work done by the battery is



Where


W is the work done by the battery


Q is the charge on the capacitor


V is the voltage across the end of the capacitor




Charge on capacitors 20μF, 30μF and 40μF are 110.76μC.


Work is done by the battery W



Question 8.

Find the charge appearing on each of he three capacitors shown in the figure.




Answer:

In the given fig. Capacitors B and C are in parallel.


The formula for parallel combination of capacitor is


Ceq=C1+C2= CA +CB= 4 + 4 =8 μF


The formula for series combination of capacitors is



From the figure, the 8 μF is connected in series with Ceqv.


Thus, we get


Putting the values in the above equation, we get




Capacitors B and C are in parallel. CC


System of B,C and A has the same capacitor values. So they exhibit the same potential difference between them.


Hence Voltage across A is =6V


The voltage across B and C is = 6V


The charge can be calculated as



Where


Q is the charge of the capacitor


C is the Capacitance of the capacitor


V is the voltage across the circuit


Charge flows through A is QA = 8×6= 48μC


Charge flows through B is QB= 4×6 =24μC


Charge flows through C is QC= 4×6 = 24μC


Charge appearing on the capacitors A, B and C is 48μC, 24μC and 24μC respectively.



Question 9.

Take C1 = 4.0 μF and C2 = 6.0 μF in figure. Calculate the equivalent capacitance of the combination between the points indicated.




Answer:

Let's name the points indicated in fig as A and B.


In a) C1 and C2 are in parallel. If we redraw the diagram that will look like









In b) also C1 and C2 are in parallel. Redraw the diagram.








Ceq=Ca+Cb=5+5=10μF



Equalent capacitance in figa) is 5μF.


Equalent capacitance in figb) is 10μF.



Question 10.

Find the charge supplied by the battery in the arrangement shown in the figure.




Answer:

Given V= 10V


C1=5


C2=6


Formula used :


Charge supplied by the battery is



Where


C is the capacitance of the capacitor


V is the voltage across the potential difference


Q is the Charge on the capacitor


First, we have to calculate the capacitance C .To do this short circuit the voltage source and open circuit the current source. We don’t have any current sources over here. So short circuit the Voltage source. Then two capacitors will come to parallel. In parallel connection of the capacitor we add the capacitor values.



Equalent Capacitance is





Charge Q can be calculated as



Charge supplied by the battery is



Question 11.

The outer cylinders of two cylindrical capacitors of capacitance 2.2 μf each are kept in contact, and the inner cylinders are connected through a wire. A battery of emf 10V is connected as shown in the figure. Find the total charge supplied by the battery to the inner cylinders.




Answer:

Lets take inner cylinders as A and B. and outer cylinders as A1 and B1. Inner cylinders A and B are connected through a wire. Outer cylinders kept in contact. If we draw the diagram, it will be look like as fig.



Given


capacitance C= 2.2μF


Voltage=10V


Capacitors are kept in parallel. So the potential difference across them is the same.


Formula used:


The magnitude of the charge on each capacitor is



Where


Q is the charge of the capacitor


C is the Capacitance of the capacitor


V is the voltage across the capacitor



Inner cylinders of the capacitor are connected to the positive terminal of the battery. So the charge on each of them is +22μC


Net charge on the inner cylinders is = 22μC+22μC= +44μC


Note: If it is asked for a charge on outer cylinders of the capacitor. A net charge will be equal to -44μC because they are connected to the negative terminal of the battery).



Question 12.

Two conducting spheres of radii R1 and R2 are kept widely separated from each other. What is their individual capacitance? If the spheres are connected by a metal wire, what will be the capacitance of the combination? Think in terms of series-parallel connections.


Answer:

The radius of conducting sphere 1 =R1


Radius conducting sphere 2 =R2


First, we need to calculate the capacitance of isolated charged sphere.


Formula used :


The capacitance of a sphere is given by the formula



Where


C is the capacitance of the capacitor


ε₀ is the permittivity of the free space is ε₀=


Taking limits as aR and b∞, Capacitance of charged sphere is found by imagining the concentric sphere with an infinite radius having some -Q charge).



The capacitance of isolated charge sphere is



The capacitance of isolated charge sphere 1 is


The capacitance of isolated charge sphere 2 is


If the two spheres are connected by a metal wire, then the charge will flow one sphere to another up to their potential becomes the same.


The potential will be the same only when they are connected in parallel. So two spheres are connected by a metal wire in parallel.





The capacitance of individual spheres of radius R1 and R2 is C1=4πε₀R1 and C2=4πε₀R2 respectively.


Combinational capacitance when charged spheres are connected by a wire is 4πε₀R1+R2).



Question 13.

Each of the capacitors shown in the figure has a capacitance of 2 μF. Find the equivalent capacitance of the assembly between the points A and B. Suppose, a battery of emf 60 volts is connected between A and B. Find the potential difference appearing on the individual capacitors.




Answer:

Given


Capacitance C= 2μF


The equalent capacitance of the first row is calculated as






The capacitance of each row is the same, and it is equal to


All the three rows are arranged in parallel. So, Voltage or potential difference across each row is the same and is equal to 60V.


So, Voltage across each capacitor is =20V.



The equalent capacitance between A and B is


Potential difference across each capacitor is 20V.



Question 14.

It is required to construct a 10 μF capacitor which can be connected across a 200V battery. Capacitors of capacitance 10 μF are available, but they can withstand the only 50V. Design a combination which can yield the desired result.


Answer:

Requirement : We have to construct a 10μF capacitor, and it has to connect across a 200V battery.


Capacitors of 10μF are available, but the voltage rating is 50V only. By using these capacitors with this voltage rating, we have to meet our requirement.


Let's assume some X capacitors are placed in series. 200V battery connected across the.


So




We have to construct 4 capacitors in a series so that we get the potential difference of 200V


If we calculate the capacitance of the parallel combination of four 10μF capacitors



Putting the value of the capacitor in the above formula, we get



Thus, the capacitance of the combination is C=2.5μF


With this arrangement, we get the required potential difference value, but we are not getting the capacitor value 10μF instead of this we get only 2.5μF. So we have to add some columns. Let us take Y as columns,




So we have to add 4 columns as the same row.


That circuit will look like



Finally, the above fig will be the design for our requirements; each capacitor value is with voltage rating 50V.



Question 15.

Take the potential of the point B in figure to be zero.

a) Find the potentials at the points C and D.

b) If a capacitor is connected between C and D, what charge will appear on this capacitor?




Answer:

From the fig. the capacitors 4μF and 8μF are in series.



Capacitors 3μF and 6μF are in series



C1 and C2 are in parallel combination


Two rows are in parallel. So the voltage across each row is the same, and that is equal to 50V.


The charge on the branch ACB is


Charge on the branch ADB is


QADB


Voltage,


Where


Q is the charge on the capacitor


C is the capacitance of the capacitor


The voltage at 4μF is =


The voltage at 8μF is =


Voltage at node C is =V


The voltage at 3μF is


The voltage at 6μF is


The voltage at node



b) if a capacitor is connected between node C and D. if we redraw the circuit, it will look like. Here bridge is balanced at the condition





the charge on the capacitor will be zero.


Because the bridge is balanced so the potential difference between C and D will be zero. So no charge flow will occur.


The voltage at node C and node D is same and is equal to


If a capacitor is connected between node C and D, the charge flow will be zero.



Question 16.

Find the equivalent capacitances of the system shown in figure between the points a and b.




Answer:

C1 and C2 are in series Equivalent capacitance,




The capacitance Ca, Cb and C3 are connected in parallel combination across each other



Putting the values in the above equation, we get





Equalent capacitance between a and b is



Question 17.

A capacitor is mad of a flat plate of area A and the second plate having a stair-like structure as shown in the figure. The width of each stair is a, and the height is b. Find the capacitance of the assembly.




Answer:

It looks like this capacitor is made up of 3 capacitors with different d separation between the plates) and arranged in parallel. Redraw the fig. it looks like



Area of the flat plate is = A


Width of the second plate is the same for all the three capacitors is =a


For C1 area is A1 = A/3


C2 area is A2 = A/3


C3 area is A3 = A/3


Height of the second plate of three capacitors is same and is =a


d1= d


d2= d+b)


d3=d +b +b)=d+2b)


Formula used:


Capacitance can be calculated by the






Capacitors are in parallel.




The capacitance of the assembly of the capacitors is




Question 18.

A cylindrical capacitor is constructed using two coaxial cylinders of the same length 10 cm and of radii 2 mm and 4 mm.

a) Calculate the capacitance.

b) Another capacitor of the same length is constructed with cylinders of radii 4 mm and 8 mm. Calculate the capacitance.


Answer:

Given


Length, l = 10cm


Radius, R1 = 2mm


Radius, R2 = 4mm



Formula used :


a) The capacitance of the cylindrical capacitor is given by



Where


l→ length of the cylinder


R2→ radius of outer cylinder


R1→ radius of inner cylinder permittivity of the free space


ε₀→ permittivity of the free space = ε₀=


Putting the value in the above formula, we get



b) Another cylindrical capacitor of same but different radius R1=4mm and R2= 8mm


If we compare the radii in a) with b), they give the same ratio




So capacitance is also same as a) is



Capacitance of cylindrical capacitor for both a) and b) is same and is =8pF



Question 19.

A 100 μF capacitor is charged to a potential difference of 24V. It is connected to an uncharged capacitor of capacitance 20 μF. What will be the new potential difference across the 100 pF capacitor?


Answer:

Given


for charged capacitor C1 =100μF


V=24V


Formula used :


Charge is given by the formula


Q=CV


Where


Q is the charge of the capacitor


C is the capacitance of the capacitor


V is the voltage across the capacitor


Putting the values in the above formula, we get



This capacitor is connected to an uncharged capacitor of C2=20μF


When a charged capacitor is connected to an uncharged capacitor, then the total charge will be equal to



The potential difference across both capacitors will be the same.






Substitute in eq1)





New potential difference is =


If 100 μF capacitor which is charged to 24V is connected to an uncharged capacitor of 20 μF then potential difference across it is 20V.



Question 20.

Each capacitor shown in the figure has a capacitance of 5.0 μF. The emf of the battery is 50V. How much charge will flow through AB if the switch S is closed?




Answer:

Given :


Capacitance C= 5.0 μF


Voltage V= 50V


Formula used :


1) If switch S is closed, it will be a short circuit. Current flow always chooses a low resistance path. No current will flow through capacitor at switch S., So we don’t need to consider it. Finally, we will left with two capacitor which are in parallel. We add the capacitance when the capacitors are in parallel.


Ceqv=C1+C2


Where


C1, C2 are the capacitance of the capacitors


2) Charge supplied by the battery



Where


C is the capacitance of the capacitor


V is the voltage across the capacitance


Q is the charge on the capacitor


Putting the values in the formula 1, we get


Ceqv= 5+5=10μF


Putting the values in the formula 1, we get



Charge supplied by the battery Q=500μC.


But when the switch has not connected the charge Q=Ceq×V


The capacitors are connected in series connection, we get




After switch S is closed the initial charge stored in the capacitor will discharge.


Note: Q1 will be negative because the capacitor is discharging.)


So the net charge flows from A to B is


Total Charge will flow through A and B when switch S is closed.



Question 21.

The particle P shown in figure has a mass of 10 mg and a charge of –0.01 μC. Each plate has a surface area 100 cm2 on one side. What potential difference V should be applied to the combination to hold the particle P in equilibrium?




Answer:

43 mV


Given,


Mass of the particle, m10 mg


The charge of the particle, q -0.01 μC -0.01 10-6 C


The capacitance of each pair of the parallel capacitor plates, C0.04 μF 0.0410-6 F


Area of each capacitor plates, A 100 cm210010-4 m2


V is the potential difference required for the particle to be in equilibrium?


Formula used


For the particle of mass ‘m’ to stay in equilibrium in the given set up, the weight of the particle W) should be opposed by the electric force F), acting on the same charged particle. The electric force is exerted by the electric field in between the capacitor plates. As the weight is acting downward, the electrical force should act upward for the equilibrium.


So,



Or



Where,


g Acceleration due to gravity 9.81m/s2


q charge of the particle -0.01 μC -0.01 10-6 C;


m10 mg10×10-4kg;


E Magnitude of Electric field in between the capacitor plates;


But from Gauss’s law, we have,



Where,


Q Charge on the capacitor plates same on both capacitors for series arrangement)


ε0Permittivity of free space 8.85 10-12Fm-1


A= Area of the plate in the parallel plate capacitor10010-4 m2


We know that, for capacitors connected in series across the voltage V, the effective capacitance, Ceff will be



Or,



Here C1=C2= C = 0.04 μF


Hence,



Or,



With this, we can calculate the value of charge stored Q) in the given capacitor arrangement as,



Where, V is the potential difference required to produce enough electric field to oppose the weight of the particle.


Putting eqn.3 in eqn.2, we get,



Now, substituting eeqn.4 in eqn.1, we get,



Or,



Substituting the known values, we get



Or,


V=0.434V=43.4 mV


Hence, to keep the particle of mass 10mg, the potential difference in the set up should be 43 mV.



Question 22.

Both the capacitors shown in figure are made of square plates of edge a. The separations between the plates of the capacitors are d1 and d2 as shown in the figure. A potential difference V is applied between the points a and b. An electron is projected between the plates of the upper capacitor along the central line. With what minimum speed should the electron be projected so that it does not collide with any plate? Consider only the electric forces.




Answer:

Given,


d1,d2 are the separations between capacitor plates in the upper and lower capacitors respectively.


a is the length of each plate


Area of each plates a2


Sy is the distance that the electron must travel in order to avoid collision in Y-direction d1/2


Sx is the distance that the electron must travel in order to avoid collision in X-direction a


V is the potential difference between the given series arrangement of capacitors.


e is the charge of electron released in between the plates


Formula used


In order to avoid a collision with plates, the electron should have an initial velocity, v. Hence, with ‘v’ velocity, the electron should travel a distance of ‘d1/2’ in Y-direction and ‘a’ in X-direction



Since the electric field is acting only in Y-direction, the electron will travel with constant velocity, v, in X-direction.


Hene the external force, neglecting gravitational and other forces, acting on the electron is the force due to the electric fieldqE). Hence, according to Newton’s second law of motion, we can write,




Where,


mmass of electron;


ay acceleration of electron in Y-direction;


q=e=charge of electron;


E= Magnitude of Electric field acting between the plates of capacitor.


We know that the distance that must be traveled in X-directiona


So, by the equations of motion, this can be represented as,



Or,



Where,


t time taken to travel ‘a’ distance


Acceleration in X-direction is Zero)


And the distance that must be traveled in Y-directiond1/2


Hence, by the equation of motion, assuming no initial velocity in Y-direction as the electron is projected horizontally.



From eqn.1 and eqn.2, eqn. 3 can be modified as,



Now, let C1 and C2 be the capacitance of the upper and lower capacitors. Hence the effective capacitance, Ceff of the series arrangement is,



Or,



Where,


and


ε0 Permittivity of free space, in between the capacitor plates.


Hence,



Or,



Now, the magnitude of electric field, E, in the upper capacitor is given by,



Where, V1 Potential difference in the upper capacitor and is equal to,



Where,


Q= charge in each capacitor total charge in the arrangement, since it is a series arrangement


Hence, Q can be calculated as,



Where V total potential difference


Or,



Substituting the above equation and the value of C1 in eqn.6, we get,



Or,



Substituting the above expression in eqn.5, we get,



Or,



Substituting the above expression in eqn.4, we get



By re-arranging,



The above expression is the least value of horizontal initial velocity needed for the electron to cross the capacitor plates without collision.



Question 23.

The plates of a capacitor are 2.00 cm apart. An electron-proton pair is released somewhere in the gap between the plates and it is found that the proton reaches the negative plate at the same time as the electron reaches the positive plate. At what distance from the negative plate was the pair released?


Answer:

Given,


The distance in between the capacitor plates 2cm


Formula used


Let t be the time, in seconds, with which proton and electron reach negative and positive charged plates respectively.


Let mp, me be the mass and qp, qe be the charge of proton and electron respectively.


ap, ae be the acceleration of proton and electron respectively, in direction of Electric field, E Let’s say Y-direction)


Let x= vertical distance traveled by proton to reach the negatively charged plate, in cm. Hence x is the distance is where we should place the electron-proton pair initially.


Hence, the distance traveled by electron 2-x) cm


The proton and electron are accelerated to the oppositely charged plates, and the expression for the respective acceleration can be written from Newton’s second law of motion.


So,



The external electric field acting on the proton The external electric field acting on the electron E


Hence, for proton of mass mp , the expression for second law of motion can be written as,



Here the term ‘qE’ represents the external force acting on the charged particle with a charge q in an electric field of magnitude E.


Similarly the expression for electron is,



From the above equations, the accelerations can be written as,



And



Hence, the distance travelled by proton in a time t seconds, x, by equations of motion



Or,



Similarly for electron, the distance traveled,



Or,



Now, to find x, the distance traveled by proton, we divide eqn.1 by eqn.2, that is,



Or,



But we know, charge of proton, charge of electron,


Hence the above expression will reduce to,



Now, mass of electron, me9.1×10-31 kg


And mass of proton, mp 1.67×10-27 kg


Substitution the above values in eqn.3, we get,



Or,



By rearranging the above expression we get,



Or,



Or,



Or,



Hence the pair should be released at a distance of 1.08×10-3 cm from the negative plate.



Question 24.

Convince yourself that parts a), b) and c) of figure are identical. Find the capacitance between the points A and B of the assembly.




Answer:

Given,


In the figure, part a), b), and c) are same.


Capacitances are 1μF,3μF,2μF,6μF and 5μF


Formula used


Each parts of the figure represents a bridge circuit. A bridge circuit is the one in which, two electrical paths are branched in parallel between the same potential difference, but are bridged by a third path, from intermediate points.


Inorder to check the balancing of the bridge circuits, the following conditions must be satisfied,



For a balanced bridge with capacitance arranged as shown in figure,



If this condition is satisfied the current through the C5 capacitor will be zero. Hence, C5 will be ineffective.


In the given figures, we have to check this condition before calculating the effective capacitance.


By comparing the above figure and the question figures, we can write,


C13 μF, C26 μF, C31 μF, C42 μF, C55 μF


And,



Or,



So, the balancing condition is satisfied, and hence, the 5 μF capacitor will be ineffective.


Thus the setup will reduce to the below form.



This is a simple capacitor combination, with two series connections connected in parallel. This can be solved in parts


Effective capacitance with C1 and C3 are,



Or,



Substituting the values of C1 and C3



Similarly on the other branch,



Or,



The above two series arrangements are arranged in parallel to each other across a potential difference. Hence their equivalent capacitance, Ceq, can be found by,



Or,



Hence, the equivalent capacitance in each of the arrangement will be 2.25μF.


Thus, for the case A), B) and C) the equivalent capacitance of the circuit remains constant.



Question 25.

Find the potential difference Va – Vb between the points a and b shown in each part of the figure.




Answer:

Explanation:


The potential difference Va – Vbcan be found out using Kirchoff’s loop rule. Kirchoffs loop rule states that, in any closed loops, the algebraic sum of voltage is equal to zero.


To solve a problem, follow some simple procedure as explained below with an example figure.



The two capacitors 1 μF and 3 μF are connected in series with the battery of V voltage.


We consider the loop and travel through it in any direction, clockwise or anti-clockwise. In the figure we choose to go in clockwise direction as shown.


Note: In the case of a DC source inside the loop, a change from –ve to +ve will be assigned as a positive potential.


Assume the total charge in the loop is q. So, as per kirchoff’s loop rule, the sum of voltages will be,



From this equation, we can find the unknown values depending on the problem.


This same principles are extended to the following problems.


a)



In the figure there are three loops: ABCabDA, ABCDA,CabDC. We goes in clockwise direction in every loops.


And assume, total charge, q is splitted into q1 and q2, since they branches in parallel.


So,



Applying kirchoff’s rule in CabDC, we get



Or,



Now, apply kirchoff’s rule in the loop ABCDA,


So,



Or,



But we know, q=q1+q2



So the above expression becomes,



Substituting eqn.1 in eqn.2, we get



Or,



Hence, the potential difference Va – Vbis,



Or,



Hence the potential difference Va – Vbis V


b) Let’s assume there a charge of q amount is in the one loop involved.



By applying Kirchoff’s loop rule, by going in clockwise direction, starting from the point a, the sum of potential difference is,



Or,



Or,



Now, we have to find the potential difference across 2μF capacitor. Since we considering Clockwise as positive direction,



Or,



Hence



Hence Va – Vbis -8V


c)



From the figure, we can see that, the either side of the terminal a-b are similar or the loops are symmetrical with respect to the terminal a-b. The charge in either of the loop will be same, which can be assumed as q.


Hence the effect on the 5 μF capacitor due to the loop on the left side will be cancelled by the loop of the right side. The charging on the 5 μF due to the left loop will get nullified by the charging by the right side loop. Hence there will be no charge accumulation on the 5 μF capacitor due to either of the battery due to their opposite orientation and symmetry.


Which means, between the terminals a-b,



Hence the Potential difference across 5μF,



Or,



Hence Va – Vbis 0V


d)


The three branches are connected in parallel across the terminal a-b.


The potential difference Va – Vbcan be found out by,



Where the net charge and net capacitance are the algebraic sum of charges and capacitance ein each branches. So,



In the upper branch, Capacitance is 4μF, and Charge, Q is,


Charge on the capacitor,


Where


C is the capacitance of the capacitor


V is the potential difference across the end of the capacitor.


Or,



In the upper branch, Capacitance is 2μF, and Charge,Q is,



Or,



In the bottom branch, Capacitance is 1μF, and Charge, Q is,



Or,



Hence Net charge between a-b, by adding all the charges,Qnet



And Net capacitance, Cnet



Hence from eqn.1.



Or,



Hence Va – Vbis 10.3V



Question 26.

Find the equivalent capacitances of the combinations shown in figure between the indicated points.




Answer:

Figure ‘a’ and ‘b’ can be solved using Y- Delta transformation while figure ‘c’ and ‘d’ can be solved using the concept of Balanced bridge circuit.


Y- Delta or Star-Delta) Transformation:


The Y-Delta transformation technique is used to simplify electrical circuits. It is an extension of Kirchoff’s Loop Rule.



As shown on the figure, the capacitance arranged in between 3 terminals of the first figure can be transformed into the form shown in the second figure. As we converts from the first form to the second one, the capacitance P,Q and R will be replaced by capacitance A, B and C.


The capacitance between terminals 1 and 2 in the second figure corresponding to that of in the first figure, can be written as,



Similarly between terminals 2 and 3 will be



Similarly between terminals 3 and 1 will be



With these values of B, C, and A , the first figure can be transformed into an easier second figure.


a)



In the figure ‘a’ , as the circuit is not balanced ∵ ), this must be changed into a simpler form using Y-Delta transformation.


Below we consider the capacitance in the ‘circled portion’, and by the transformation equations,



The capacitance equivalent to 1μF and 3μF is,



Similarly, corresponding to the capacitance 1μF and 4μF, the equivalent capacitance after transformation is,



Similarly, corresponding to the capacitance 3μF and 4μF, the equivalent capacitance after transformation is,



Hence the resultant figure can be drawn as shown,



All the values are in μF)



And it can be further simplified, by re-arranging parallel and series arrangements as shown in figure below.


The above arrangement of capacitances is a simple one, and can be done using the basic equations.


First, consider the two parallel arrangements at the bottom, the equivalent capacitance in the left one is,



Or,



Similarly for the bottom right arrangement,



Or,



Hence the effective capacitance, considering two series capacitance Ceq1,Ceq2) connected in series with the 3/8 μF, is




Hence, the Effective capacitance between the terminals is 11/6)μF


b)


In figure ‘b’ we have to apply Y-Delta transformation at two portions, as circled in the picture below.




We apply Y- Delta transformation in each circled portion.


In the upper portion,


The capacitance equivalent to 1μF and 3μF is,



Similarly, corresponding to the capacitance 1μF and 4μF, the equivalent capacitance after transformation is,



Similarly, corresponding to the capacitance 3μF and 4μF, the equivalent capacitance after transformation is,



At the lower circled portion,


The same values will come , as the two portions are symmetrical with respect to the central horizontal line. Hence the arrangement becomes,




All the values are in μF)



By simplifying further, it becomes,



Hence Effective capacitance is,




Hence, the Effective capacitance between the terminals is 11/4)μF.


c)



This problem can be done by either Y-Delta transformation or by the concept of balanced bridge circuits. Here we choose the concept of balanced bridge circuits for simplicity.



In the below figure, the circled portion is a balance bridge since it obeys balancing condition which is,



Or,



And hence the 5μF capacitor will be ineffective as per the principle. Hence the arrangement will be reduced into,



Or, by combining the series capacitance together, it will be reduced into,



This is a simple parallel arrangement, and effective capacitance can be calculated as,



By substituting the values, we get




Hence, the Effective capacitance between the terminals is 8μF.


d)


This problem can be done by the concept of balanced bridge circuits. There are three balanced bridges present in the arrangement.


For which


Since,



and



They are balanced and hence the three 6 μF capacitance will be ineffective.



Hence the resultant arrangement will be,


It is further reduced, by combining series capacitors together, into,



This is a simple parallel arrangement, and effective capacitance can be calculated as,



Or,




Hence, the Effective capacitance between the terminals is 8μF.



Question 27.

Find the capacitance of the combination shown in figure between A and B.




Answer:

The question figure is a simple arrangement of parallel andseries configurations.


Let us number each capacitor as C1, C2,… and C8 for simplification.



In the figure 5th and 1st capacitors are in series, hence the effective capacitance, C51 is



Or,




Now, C51 and C6 are in parallel,



Hence the effective capacitance, C61 is,



On substituting,



Now, C61 and C2 are in series, hence the effective capacitance, C62 is,



Or,




This above pattern repeats for 2 more times. Hence at the end, the effective capacitance, Ceff will be 1μF,


So,


The capacitance of the combination is hence 1μF.



Question 28.

Find the equivalent capacitance of the infinite ladder shown in figure between the points A and B.




Answer:

This is an infinite series and hence deletion or addition of any repetitive portions of the arrangement does not affect the overall effect. So, let’s convert this into a simpler figure for calculation.



In the above figure, ‘C’ represents the effective capacitance of the infinite ladder. For the calculations, we have added a 1μF and a 2μF as shown since they both constitute the repetitive portion of the question figure. Hence C and 2μF are in series and they instead is parallel to 1μF.


Now, we calculate the value of C as,



Which is equals to C itself,


So,


Or,



Or,



Or,



Or,



Or,



Since capacitance value cannot be negative, we neglect C=-1μF. Hence the equivalent capacitance of the infinite ladder is 2μF.



Question 29.

A finite ladder is constructed by connecting several sections of 2 μF, 4 μF capacitor combinations as shown in figure. It is terminated by a capacitor of capacitance C. What value should be chosen for C, such that the equivalent capacitance of the ladder between the points A and B becomes independent of the number of sections in between?




Answer:

Given,


The capacitance C should be equal to the equivalent capacitance.


Since the arrangement is an infinite series, addition or deletion of the repetiting components which is the 2 μF, 4 μF capacitor combinations) would not make any effect on the overall capacitance.



Hence, for simplification, we represent it as shown below,


In the figure , C in μF) represents the capacitance that gives the same value for equivalent capacitance to the infinite ladder even after it is terminated at the end. So that C and 4 μF are in series, and these are parallel to 2μF.


In this case, the effective capacitance Ceff



Which is equals to C itself, since C should not alter the effective capacitance.


So,


Or,



Or,



Or,



Or,




Since capacitance value cannot be negative, we neglect C=-2μF. Hence the equivalent capacitance of the infinite ladder is 4μF.


So, the value of capacitance that should be assigned with the terminating capacitor is 4 μF.



Question 30.

A charge of +2.0 × 10–8 C is placed on the positive plate and a charge of –1.0 × 10–8 C on the negative plate of a parallel-plate capacitor of capacitance 1.2 × 10–3 μF. Calculate the potential difference developed between the plates.


Answer:

Given,


Charge on plate 1, Q1 = +2.0 × 10–8C


Charge on plate 2, Q2 = –1.0 × 10–8C


Capacitance of the capacitor, C= 1.2 × 10–3 μF= 1.2 × 10–9 F


Formula used


We know that for a capacitor with net charge, Q and capacitance, C, the Potential difference deceloped in between the plates, V is,



The charges on the inner plates of the capacitor with plates having charges Q1 and Q2 is,



Note: Charges on the outer plates of the capacitor with plates having charges Q1 and Q2 is,



In the given example, the plates has individual charges Q1 and Q2. But when they placed as a capacitor, their charges re-arrange and equal and opposite charges will be distributed in each plates. The total net charge, Qnet on the inner sides of each plates will be



Substituting the values,




Hence the inner side of each plates will have a charge of ±1.5 × 10–8C.


Hence from eqn.1, Potential difference, V is



Or,



Hence the potential difference developed between the plates is 12.5V



Question 31.

A charge of 20 μC is placed on the positive plate of an isolated parallel-plate capacitor of capacitance 10 μF. Calculate the potential difference developed between the plates.


Answer:

Explanation:


Given,


Charge on plate 1, Q1 = 20 μC


Charge on plate 2, Q2 = 0C Since no charge is given to the other plate and the setup is isolated)


Capacitance of the capacitor, C= 10 μF


Formula used


We know that for a capacitor with net charge, Q and capacitance, C, the Potential difference deceloped in between the plates, V is,



The charges on the inner plates of the capacitor with plates having charges Q1 and Q2 is,



Note: Charges on the outer plates of the capacitor with plates having charges Q1 and Q2 is,



In the given question, the charges on the inner plates , according to above formulas,



Hence from eqn.1, the potential difference



Hence the potential difference developed in between the plates is 1V.



Question 32.

A charge of 1 μC is given to one plate of a parallel-plate capacitor of capacitance 0.1 μF and a charge of 2 μC is given to the other plate. Find the potential difference developed between the plates.


Answer:

Explanation:


Given,


Charge on plate 1, Q1 = 1 μC


Charge on plate 2, Q2 = 2 μC


Capacitance of the capacitor, C= 0.1 μF


Formula used


We know that for a capacitor with net charge, Q and capacitance, C, the Potential difference deceloped in between the plates, V is,



The charges on the inner plates of the capacitor with plates having charges Q1 and Q2 is,



So, the charge, Q by substituting the given values, is



Hence from eqn.1, the potential difference



Hence the potential difference developed in between the plates is 5V.



Question 33.

Each of the plates shown in figure has surface area 96/ϵ0) × 10–12 Fm on one side and the separation between the consecutive plates is 4.0 mm. The emf of the battery connected is 10 volts. Find the magnitude of the charge supplied by the battery to each of the plates connected to it.




Answer:

Given,


The area of the capacitor plates, A 96/ϵ0) × 10–12 Fm


The distance in between each pairs of plates, d 4mm410-3 m


The emf of the connected battery, V 10V


Formula used


The schematic representation of distribution of charges when connected to the DC battery is shown in the figure. The plate 2) connected to the positive terminal will be positively charged and the one 4) connected to the negative terminal will be negatively charged.



The oposite charges will be induced in plates 1) and 3), whe the battery is connected as shown.


So, there will be three capacitors that are formed namely, 1-2, 2-3 and 3-4. And they are connected in series arrangement.


Let’s assume that each capacitors has a charge Q, and since they are connected in series, the total charge will also be Q.


Hence the charge, Q



Where,


V Potential difference 10V


Ceq Equivalent capacitance of the arrangement.


Now, for series arrangement, we know



And C1, C2 and C3 are the capacitance of capacitors formed by plates 1-2, 2-3 and 3-4 respectively.


Since area and the separation of all the plates are same,



And we know,


Capacitance of the capacitor,


Where


εo is the permittivity of the free space


A is the area of the plates of the capacitor


d is the separation between the plates of the capacitor


Substituting the given values in the above equation, we get




Hence, Equivalent capacitance is,



Or,



or,



Or,



Hence, from eqn.1, the charge on each pairs will be,



This is the charge on each side of the plates constituting a capacitor. But the plates connected to the battery has either positive charge or negative charge on both sides, as shown in figure. So, the total charge accumulated in the plates connected to the battery will be two times the above value.


Hence, charge on the plates connected to battery will be 2Q,



Or,



Hence the charge on the specific plates will be ±0.16μC, since one plate is positively charged and the other is negatively charged.



Question 34.

The capacitance between the adjacent plates shown in figure is 50 nF. A charge of 1.0 μC is placed on the middle plate.

a) What will be the charge on the outer surface of the upper plate?

b) Find the potential difference developed between the upper and the middle plates.




Answer:

Given,


The charge given to the middle plate Q) is 1.0 μC


The capacitance between the plates, C is 50 nF=50× 10–3 μF


Formula used


For a capacitor with net charge, Q and capacitance, C, the Potential difference deceloped in between the plates, V is,



The charges on the inner plates of the capacitor with plates having charges Q1 and Q2 is,



Charges on the outer plates of the capacitor with plates having charges Q1 and Q2 is,



a)


The charge given to the plate Q will be distributed equally on the either sides of plates as shown in figure. Hence the upper and lower sides of plate Q will be charged to +0.5 μC.



Here, we get two capacitors namingly as P-Q and Q-R.


In capacitor P-Q, the upper plate is neither connected to any battery nor given any charges. So the total charge on the plate is 0C. But when it is made into a capacitor plate, a charge is induced in it from the plate Q. The amount of the charge can be calculated from the eqn.2,


Which is,



Where Q1 is the charge on one plate Q= 1.0 μC


And Q2 is the charge on plate P = 0C


Hence by substituting in the above equation, we get,



Hence the inner surfaces get a charge of ±0.5μC on each plates. Since the plate Q is positively charged, Plate P will get -0.5μC charge.


But we know that the net charge on plate P is zero. Hence to nutralise the inner surface charge, the outer surface will get a charge of +0.5μC


b) From the above calculation, we found that the inner surfaces of the capacitor P-Q has a charge of ±0.5μC.


The potential difference between the plates can be found by the eqn.1, as



Hence the potential difference between the upper and middle plates of the arrangements is 10V.



Question 35.

Consider the situation of the previous problem. If 1.0 μC is placed on the upper plate instead of the middle, what will be the potential difference between

a) the upper and the middle plates and

b) the middle and the lower plates?


Answer:

Given,


Charge given to the upper plate, plate P, is 1.0 μC.


Formula used


For a capacitor with net charge, Q and capacitance, C, the Potential difference deceloped in between the plates, V is,



The charges on the inner plates of the capacitor with plates having charges Q1 and Q2 is,



a)


By giving a charge of 1.0 μC to plate P, it will get distributed on either side of the plate as +0.5 μC. The other plates get induced with this charge as shown in figure.




To find the charge on the plate Q, eqn.2 shall be used. So we get,



Where Q1 is the charge on one plate P= 1.0 μC


And Q2 is the charge on plate Q = 0C


Hence by substituting in the above equation, we get,



So the upper face of plate Q will get a charge of -0.5 μC and this will induce a charge of +0.5 μC on the bottom side of plate Q.


Hence the potential difference in capacitor P-Q, by eqn.1 is



Where Q= 0.5 μC


And C= 50×10-3μF


Hence,



So the potential difference in between the upper and middle plates is 10V


b)


From the above condition, the upper face of plate Q will get a charge of -0.5 μC and this will induce a charge of +0.5 μC on the bottom side of plate Q.


We know that, the capacitor Q-R is made of the bottom surface of plate Q and the upper side of plate R. As the bottom surface of plate Q already has a charge of +0.5 μC, it will induce -0.5 μC charge on the upper face of plate R As shown in figure).


Hence the potential difference in between the lower and middle plates can be calculated from the eqn.1, as



Where Q is the charge in each plates=±0.5 μC


And C= 50×10-3μF


Hence,



So the potential difference in between the middle and lower plates is 10V



Question 36.

Two capacitors of capacitances 20.0 pF and 50.0 pF are connected in series with a 6.00V battery. Find

a) the potential difference across each capacitor and

b) the energy stored in each capacitor.


Answer:

Given,


Capacitances of the two capacitors, are 20.0 pF and 50.0 pF.


The voltage of the battery, V is 6V


Formula used


Let us represent the arrangement as



In a series arrangement the the charge on both the capacitance are same equal to total charge),can be found out by the equation,



Where Q and V represents the Charge and Potential difference respectively.


in series arrangement with Capacitance C1 and C2, Ceff can be found out as,



Or,



And thus the potential difference on each capacitance, V1 and V2 can be calculated by the below relations,



And,



Now,


The energy stored in a capacitor,E in Jules) can be found out by the relation,



Where


C is the capacitance of the capacitor in Farad


V is the potential difference across the capacitor.


a) First we calculate the ewuivalent capacitance by eqn.2.



Where C120 pF and C2=50pF


So,


Hence, the total charge, Q from eqn.1 is



To find potential difference on each capacitor, we use eqn.3 and eqn.4. So the potential difference on 50pF capacitor is,



Or,



Similarly, on 20pF capacitor, V2 is



Or,



Hence the potential differences across 50pF and 20pF capacitors are 1.714V and 4.29V respectively.


b) Energy stored in each capacitors can be calculat4ed by eqn.5


Hence for, 20pF capacitance across 4.29V potential difference, energy stored is,



Similarly for, 50pF capacitance across 1.71V potential difference, energy stored is,



Hence Energy stored in each capacitors are 73.5pJ and 184.04pJ for 50pF and 20pF capacitors respectively.



Question 37.

Two capacitors of capacitances 4.0 μF and 6.0 μF are connected in series with a battery of 20V. Find the energy supplied by the battery.


Answer:

Given,


Capacitance are 4.0 μF C1) and 6.0 μF C2)


The voltage of battery, V is 20V


Formula used


Let us represent the arrangement as



In a series arrangement the the charge on both the capacitance are same, and can be found out by the equation,



Where Q and V represents the Charge and Potential difference respectively.


in series arrangement with Capacitance C1 and C2, Ceff can be found out as,



Or,



The energy stored in the capacitor,E in Jules) can be found out by the relation,



Where C is the capacitance of the capacitor in Farad and V is the potential difference across the capacitor.


We have to find the equivalent capacitance by eqn.2


So,



Or,



Now the energy supplied by the battery is equivalent to the energy stored in the equivalent capacitor with capacitance Ceff. Hence by eqn.3, we get



Or,



The supplied energy will be twice of the stored energy, since half of the supplied energy will be dissipated by the resistance of the circuit.


Hence the supplied energy will be



Hence an amount of 960 μJ will be supplied by the battery.



Question 38.

Each capacitor in figure has a capacitance of 10 μF. The emf of the battery is 100V. Find the energy stored in each of the four capacitors.




Answer:

Given, capacitance of a, b, c, d capacitors are 10 μF each.


The voltage of the DC battery is 100V


Formula used


Energy stored in a capacitor can be calculated from the relation,



Or,



Where C represents the capacitance, V is the potential difference across the capacitor and Q is the charge in the capacitor.


And, effective capacitance of capacitors C1 and C2 arranged in series is



And those connected in parallel is


Considering three capacitors)


The capacitors b and c are in parallel. For simplification, we reduce it into capacitor bc as shown,



and the capacitance of bc is, from eqn.4



By substituting the values,



Now the whole arrangement is a series connection and charges in each capacitor will be the same.


To find out effective capacitance of this arrangement, we find equivalent capacitance, Cad between a and d initially, by eqn.3



By substituting the values,



Now the total capacitance considering Cadand Cbc in series, using eqn.3,


Or,



The capacitors a ,d and the parallel arrangement will have same charge,Q in it, which can be calculated as,



Where,


Ceff= Capacitance, V= Potential difference=100V


By substitution, we get , Q as



The energy stored in a and d are same due to the same capacitance value and the same charge accumulation. So energy stored in a and d are, from eqn.2



Or,



In the parallel arrangement, the charge, Q=400μC will be splitted in half as the two branches are symmetrical. So each capacitors b and c will have Q=200μC amount of charge.


Hence, by the energy relation, eqn.2 , the energy in each capacitors b and c, will be,



Or,



Hence 8mJ will be stored in the capacitors a and d , while 2mJ will be stored in b and c.



Question 39.

A capacitor with stored energy 4.0 J is connected with an identical capacitor with no identical capacitor with no electric field in between. Find the total energy stored in the two capacitors.


Answer:

Given,


The stored energy in the first capacitor is 4.0J


Formula used


Energy stored in a capacitor of capacitance C across a potential difference V is,


Energy stored in the capacitor,


Whenever an uncharged capacitor is connected with a charged capacitor, the charge will redistribute according to the capacitance of both of the capacitors.


In the given case, both the capacitors are identical and hence the charge will distribute equally in both.


We assume that the charge in the first capacitor is initially as q. After the charge distribution, the charge on both capacitors will be q/2.


Since the capacitance are equal and there is no electric field placed in between, according to the eqn.1 the energy stored in both the capacitors are same.


So, by conservation of energy, the total 4J will be distributed to both of the capacitors. So each capacitor will store energy of amount 2J.



Question 40.

A capacitor of capacitance 2.0 μF is charged to a potential difference of 12V. It is then connected to an uncharged capacitor of capacitance 4.0 μF as shown in figure. Find

a) the charge on each of the two capacitors after the connection,

b) the electrostatic energy stored in each of the two capacitors and

c) the heat produced during the charge transfer from use capacitor to the other.




Answer:

Given,


Potential difference, V is 12V


Capacitance of initially charged capacitor, C1 is 2 μF


Capacitance of initially uncharged capacitor, C2 is 4 μF


Formula used,


Energy stored in a capacitor of capacitance C and charge Q is,



Energy stored in a capacitor of capacitance C across a potential difference V is,



a)


Initial charge on C1capacitor, Q1 is



Or,



Now, let’s assume that after connecting the second capacitor C2, the charge on C1 and C2 as q1 and q2 respectively.


So,


We know that for a parallel arrangement of capacitors across a single battery, the potential differences are the same. So



Or, by substituting the values for C1 and C2, we can re-write it as,



Or,



Substituting eqn.3 in eqn.2, we get



Or,



And



b) Energy stored on each capacitor, by eqn.1 is


On C1



Or,



Similarly,


On C2



Or,



Hence the energy stored is 16μJ and 32μJ on 2μF and 4μF capacitors respectively.


c) For heat dissipation, we have to find the initial energy stored.


Hence from eqn.1, the initial energy with 2μF capacitor only in the circuit, Eb is



Where V=12V


So after substitution,



Hence heat produced is the difference between the initial energy and the algebraic sum of the energy stored after connection.


So,



Or,



The heat produced/dissipated during the charging is 96μJ



Question 41.

A point charge Q is placed at the origin. Find the electrostatic energy stored outside the sphere of radius R centred at the origin.


Answer:

Given,


R is the radius of the sphere and Q is a point charge


Formula used


We know that for a sphere or a point charge, the capacitance can be found out by the equation,



Now, to find energy stored, we have the relation,



Here the point charge has Q amount of charge and capacitance C is as given above. So by substitution,



Hence the expression for energy stored on a sphere around a point charge placed at the origin is Q2/8πε0×R) J



Question 42.

A metal sphere of radius R is charged to a potential V.

a) Find the electrostatic energy stored in the electric field within a concentric sphere of radius 2R.

b) Show that the electrostatic field energy stored outside the sphere of radius 2R equals that stored within it.


Answer:

J

Explanation:


The given condition is represented in the figure. The outer sphere has a radius 2R while the metal sphere has a radius R.



Now potential difference, V of the sphere is given by,



Where Q and C represents Charge and Capacitance of sphere


For sphere of radius R, C is



Substituting this in eqn.1, we get,



a)


Energy density at a distance r from the centre is,



Or



Consider a spherical element at a distance r from the centre, with a thickness dr, such that R>r>2R.


Now the volume of the spherical element is,



So, energy stored will be



Or



For finding the electrostatic energy on a surface at 2R, we have to integrate the expression for dUE in between R and 2R. So,



Or,



But from eqn.2,



Hence, UE becomes,



Or,


Electrical energy at a distance 2R is



b)


To find the electrostatic stored energy outside the radius 2R, we integrate the above expression for differential of stored energy from 2R to infinity.


So,



Or,



By substituting q as 4πε0×R×V in the above expression, we get,



Or it will reduce to,



This is same as that of inside the sphere of radius 2R.


Thus electrostatic field energy stored outside the sphere of radius 2R equals that stored within it.



Question 43.

A large conducting plane has a surface charge density 1.0 × 10–4 C m–2. Find the electrostatic energy stored in a cubical volume of edge 1.0 cm in front of the plane.


Answer:

Given,


Surface charge density,σ1.0 × 10–4 C m–2


Edge length of the cube, e=1.0 cm=0.01m


Permittivity of free space, ε0= 8.85×10-12 F/m


Formula used


For a conducting plate infinite length), the electric field, E is,



And the electrostatic energy density or the energy per volume is,



Substituting eqn.1 in eqn.2 will result in,



Now the energy stored in volume V is



In the problem, we have to find the force inside a cube of edge e length.


So, we replace V with e3 in eqn.3


So,



Now, substituting the known values in the above equation, it becomes,



Or,





Question 44.

A parallel-plate capacitor having plate area 20 cm2 and separation between the plates 1.00 mm is connected to a battery of 12.0 V. The plates are pulled apart to increase the separation to 2.0 mm.

a) Calculate the charge flown through the circuit during the process.

b) How much energy is absorbed by the battery during the process?

c) Calculate the stored energy in the electric field before and after the process.

d) Using the expression for the force between the plates, find the work done by the person pulling the plates apart.

e) Show and justify that no heat is produced during this transfer of charge as the separation is increased.


Answer:

Given –


Plate area 20 cm2 = 0.002m2


Separation between the plates 1.00 mm = 0.001m


Battery Voltage = 12.0 V


We know capacitance, C


1)



Where,


A= Plate Area


d= separation between the plates,


0 = Permittivity of free space


= 8.854 × 10-12 m-3 kg-1 s4 A2


From1),


Capacitance when distance d = 0.001m, C1



Substituting values,


= 2)


When The plates are pulled apart to increase the separation to –


2.0 mm = 0.002m, then capacitance C2 becomes,



Substituting values


= 3)


From 1) and 2)


4)


We know capacitance in terms of voltage is given by –


Q = CV 5)


Where


Q= charge stored on the capacitor


C= capacitance of the capacitor


V = applied voltage


Given applied v = 12V


From 2) and 3) and 5)



=


= 6)


And



7)


a) The charge flown through the circuit during the process –



From 6) and 7)



=


=


= 8)


b)Energy absorbed by the battery during the process-


We know Energy E is given by -


E=QV 9)


Where


Q = charged present on the surface


V= Applied voltage


From 8),



Applied voltage V = 12V


From 9),


Energy absorbed,



=


c)Stored energy in the electric field before and after the process


We know that stored energy in the electric field,



Before process, the energy stored -



From 2),



10)


From 3), After process, the energy stored will become



11)


d) The work done by the person pulling the plates apart.


We know, work done, W


12)


Where, f = force


d = displacement


We, know in parallel plate capacitor, the force between the plates is given by.


13)


Where, Q = charge enclosed,


σ = surface charge density,


σ, surface charge density is given by ,



From 12) and 13)


Work done,


Given, Plate area 20 cm2 = 0.002m2


Separation between the plates changed to, 2.00 mm = 0.002 m by


pulling apart.


Substituting values, from 7),







Question 45.

A capacitor having a capacitance of 100 μF is charged to a potential difference of 24V. The charging battery is disconnected and the capacitor is connected to another battery of emf 12V with the positive plate of the capacitor joined with the positive terminal of the battery.

a) Find the charges on the capacitor before and after the reconnection.

b) Find the charge flown through the 12V battery.

c) Is work done by the battery or is it done on the battery? Find its magnitude.

d) Find the decrease in electrostatic field energy.

e) Find the heat developed during the flow of charge after reconnection.


Answer:

Given



Capacitance =100 μF


Initial battery voltage used = 24V


Second voltage used = 12V


a) Charges on the capacitor before and after the reconnection.


Before reconnection, the battery used is 24V, hence




We know charge present on a capacitor is given by


q = CV 1)


Where q = charge


C=capacitance


V=voltage


Substituting in 1)






Similarly, after connection of 12V battery –


When C = 100μf


V = 12V


Using equation 1)





b) Charge flown through the 12V battery.


C = 100, V = 12V


From 1),



Substituting the values,





c) Work is done by the battery, and its magnitude is as follows


We know



Where V = applied voltage across the capacitor


W = work done and


q = charge on the surface of the parallel plate capacitor


Which gives,





is the amount of work done on the battery.


d) Decrease in electrostatic field energy


Electrostatic field energy stored is given by –



, c = capacitance


v= voltage across capacitor


Initially, electrostatic field energy stored is given by -



Final Electrostatic field energy



Decrease in Electrostatic field energy


=


=


=


Substituting the values


Capacitance =100 μF


Initial battery voltage used = 24V


Second voltage used = 12V


=



e) Heat developed during the flow of charge after reconnection


After reconnection


C = 100μc, V = 12v


We know the energy stored, E in capacitor is given by



Where c is the capacitance and v is the applied voltage


Substituting values –




This is the amount of energy developed as heat when the charge flows through the capacitor.



Question 46.

Consider the situation shown in figure. The switch S is open for a long time and then closed.

a) Find the charge flown through the battery when the switch S is closed.

b) Find the work done by the battery.

c) Find the change in energy stored in the capacitors.

d) Find the heat developed in the system.




Answer:

Given circuit as shown below -



a) Charge flown through the battery when the switch S is closed. Since the switch was open for a long time, hence the charge flown must be due to the both.


When the switch is closed, the capacitor is in series, the equivalent capacitance is given by



Now, we know the relation between capacitance, charge q and voltage v given by ,





b) Work done by the battery


We know, work done is given by



Where q is charge stored and v is the applied voltage


We have, and


Substituting the values, we get,



=


c) Change in energy stored in the capacitors


Energy stored in capacitor is given by –



Initially, the energy stored in the capacitor is given by


1)


After closing the switch, the capacitance changes to



Energy stored after closing the switch is given by -



2)


From 1) and 2)


Change in energy stored in the capacitors



=


= 3)


d) Heat developed in the system


The net change in the stored energy is wasted as heat developed in the system,


Hence, heat developed in the systems is given as-


H=∆E


Where, H is the heat developed and ∆E is the change in the stored energy in the capacitor


From 3)




Question 47.

A capacitor of capacitance 5.00 μF is charged to 24.0V and another capacitor of capacitance 6.0 μF is charged to 12.0V.

a) Find the energy stored in each capacitor.

b) The positive plate of the first capacitor is now connected to the negative plate of the second nd vice versa. Find the new charges on the capacitors.

c) Find the loss of electrostatic energy during the process.

d) Where does this energy go?


Answer:


Given:


C1=5 μF


V1=24 V


To calculate the charge present on the capacitor, we use the formula



where,


c = capacitance of the capacitor and


v = voltage across the capacitor


For first capacitor, the stored charge q1 is given by





Similarly for second capacitor, the stored charge q2 is given by-



Given, C2=6 μF and V2=12




a) Energy stored in each capacitor-


Energy stored in a capacitor is given by



Where, v = applied voltage


C =capacitance


For capacitor C1, energy stored is given by



=



Similarly, for capacitor C2, energy stored is given by





b) New charges on the capacitors when the positive plate of the first capacitor is now connected to the negative plate of the second nd vice versa



The capacitors are connected as shown on the right hand side.


The positive of first capacitor is connected to the negative of the second capacitor.


So charge flows from positive of first capacitor to the negative of the second capacitor.


Then, the net charge for connected capacitors becomes





Now, let V be the common potential of the two capacitors


Since, charge is conserved, we know that electric charge can neither be created nor be destroyed, hence net charge is always conserved.



From the conservation of charge before and after connecting, we get, common voltage V





We know,



where v = applied voltage and C is the capacitance


Using above relation, the new charges becomes-








c) Loss of electrostatic energy during the process


Energy stored in a capacitor is given by


1)


Where, v = applied voltage


C =capacitance


For capacitor C1, energy stored is given by



=


Similarly, for capacitor C2 , energy stored is given by





=


2)


But given


for c1, actual V1 = 24V


and c2, actualV2 = 12V


Using 1)



And



Now, change in energy,



3)


From 2) and3)


Loss of electrostatic energy =



d) This energy, which is lost as electrostatic energy gets converted and dissipated from the capacitor in the from of heat energy.



Question 48.

A 5.0 μF capacitor is charged to 12V. The positive plate of this capacitor is now connected to the negative terminal of a 12V battery and vice versa. Calculate the heat developed in the connecting wires


Answer:


Given


Initially 5.0 μF capacitor is charged to 12V as shown in fig.



Next, the positive plate of this capacitor is now connected to the negative terminal of a 12V battery as shown in fig.


When the capacitor is connected to the battery of 12V with first plate to positive and second plate to negative, a positive charge Q = CV appears on one plate where, C is the capacitance and v is the voltage applied, and –Q charge appears on the other.


When the polarity is reversed, a charge –Q appears on the first plate and +Q on the second plate.


A total charge of 2Q accumulates on the negative plate.



Therefore, 2Q charge passes through the battery from the negative to the positive terminal.


The battery does a work-



Where,


Q = charge and v= applied voltage


Since, a total charge of 2Q accumulates on the negative plate



We know, Q = C×v


Where c= capacitance and v= applied voltage



The energy stored in the capacitor is the same in the two cases


and the work done by battery dissipates as heat in the connecting wires.


Hence, the heat produced is -



Given , C = 5.0 μF and voltage v = 12V






Question 49.

The two square faces of a rectangular dielectric slab dielectric constant 4.0) of dimensions 20 cm × 20 cm × 1.0 mm are metal-coated. Find the capacitance between the coated surfaces.


Answer:

Given,

dielectric constant of slab = 4.0


Area of slab = 20 cm × 20 cm


Separation between slab, the thickness of the slab= 1.0 mm


We know capacitance is given by



Where


A= area of cross section


K = dielectric constant


d= separation between the slab and


0 = permittivity of free space = 8.85×10-12


Hence, capacitance is given by-





Question 50.

If the above capacitor is connected across a 6.0V battery, find

a) the charge supplied by the battery,

b) the induced charge on the dielectric and

c) the net charge appearing on one of the coated surfaces.


Answer:

Given-



Capacitance of the capacitor, C = 1.42 nF
Dielectric constant, k = 4
Voltage of the battery connected, V = 6 V


a)The charge supplied by the battery is given by-


We know from definition of capacitance, charge q on capacitor is given by -


q=CV


Where C= capacitance and V = applied voltage.




1)


b) The charge induced on the dielectric –


We know, the induced polarization charge on a dielectric material is given by-



Where,


qp = polarized charge


q = charge on the capacitance


k = dielectric constant




2)


c)The net charge appearing on one of the coated plates –


The net charge appearing will be the charge on the plat minus the charge on dielectric material


From 1) and 2)


=



Question 51.

The separation between the plates of a parallel-plate capacitor is 0.500 cm and its plate area is 100 cm2. A 0.400 cm thick metal plate is inserted into the gap with its faces parallel to the plates. Show that the capacitance of the assembly is independent of the position of the metal plate within the gap and find its value.


Answer:

Given-


Area of the plate= 100 cm2 = 0.01m


Separation between the plates = 0.500 cm = 5 × 10-3 m


Thickness of the metal, t = 4 × 10-3 m


We know capacitance is given by



Where


A= area of cross section


K = dielectric constant


d= separation between the slab and


0 = permittivity of free space = 8.85×10-12


t = Thickness of the metal


Since, it’s a metal, for metals k = infinite


Hence, capacitance is given by-




=


Here capacitance is a constant value, hence the capacitance


is independent of the position of the metal.


At any position, the net separation is dt).



Question 52.

A capacitor stores 50 μC charge when connected across a battery. When the gap between the plates is filled with a dielectric, a charge of 100 μC flows through the battery. Find the dielectric constant of the material inserted.


Answer:

Given-


Initially, the charge on the capacitor = 50 μC


Now, let the dielectric constant of the material inserted in the gap be k.


When this dielectric material is inserted, 100 μC of extra charge flows through the battery


Therefore , the net charge on the capacitor becomes


50 + 100 = 150 μC


Now, we know capacitance of a material is given by –



Where q is charge on the capacitance and v is the applied voltage


Also



Where A is the plate area and ∈0 is the permittivity of the free space.


Initially, without dielectric material inserted, capacitance is given by



1)


Similarly, with the dielectric material place, capacitance is given by



2)


On dividing 1) by 2), we get





Thus, the dielectric constant of the given material is 3.



Question 53.

A parallel-plate capacitor of capacitance 5 μF is connected to a battery of emf 6V. The separation between the plates is 2 mm.

a) Find the charge on the positive plate.

b) Find the electric field between the plates.

c) A dielectric slab of thickness 1 mm and dielectric constant 5 is inserted into the gap to occupy the lower half of it. Find the capacitance of the new combination.

d) How much charge has flown through the battery after the slab is inserted?


Answer:

Given-


Capacitance C=5 μF = F


Voltage, V=6v


Separation between plates, d=2 mm=2×10-3 m


a)The charge on the positive plate is calculated using


q = c×v


where, c is the capacitance


and v = voltage applied


Thus,


q=5 μF×6 V


=30 μC


b) The electric field between the plates of the capacitor is given by



Where, v is the applied voltage and d is the distance between the capacitor plates



=3×103 V/m


c)Given-


Distance between the plates of the capacitor, d =2×10-3 m
Dielectric constant of the dielectric material inserted, k = 5
Thickness of the dielectric material inserted, t = 1×10-3 m


capacitance of the capacitor= 5 μF


Now,
To calculate area of the plates of the capacitor,



Where,


A = area


k = dielectric constant of the material placed


d= separation between the plates


substituting the values,





When the dielectric placed in it, the capacitance becomes



t=thickness of the material


substituting the values,




1)


d)
The charge stored in the capacitor initially is -


C=5×10-6 F


Also, V=6 V


Now, we know



where v is the applied voltage and c is the capacitance




Now, when the dielectric slab is inserted ,charge on the capacitor, from 1)



Charge, Q’





Now, charge flow is given by,




Question 54.

A parallel-plate capacitor has plate area 100 cm2 and plate separation 1.0 cm. A glass plate dielectric constant 6.0) of thickness 6.0 mm and an ebonite plate dielectric constant 4.0) are inserted one over the other to fill the space between the plates of the capacitor. Find the new capacitance.


Answer:

Given:



Area of the plate, A is 100 cm2


Separation of the plate, d is 1 cm


Dielectric constant of the glass plate is 6


Thickness of the glass plate is 6.0 mm


Dielectric constant of an ebonite plate is 4.0


The given system of the capacitor will connected as shown in the fig.


The capacitors behave as two capacitors connected in series.


Let the capacitances be C1 and C2.


Now,


capacitance c



Where, A = area


k = dielectric constant of the material placed


d=separation between the plates


and is permittivity of free space whose value is


Now, first capacitor C1


1)


and


2)


Hence, the net capacitance for a series connected capacitor is given by-


3)


from 1),2), and 3)








Question 55.

A parallel-plate capacitor having plate area 400 cm2 and separation between the plates 1.0 mm is connected to a power supply of 100V. A dielectric slab of thickness 1.0 mm and dielectric constant 5.0 is inserted into the gap.

a) Find the increase in electrostatic energy.

b) If the power supply is now disconnected and the dielectric slab is taken out, find the further increase in energy.

c) Why does the energy increase in inserting the slab as well as in taking it out?


Answer:

Given,


Area, A = 400cm2 = 400 × 10–4m2


distance between plates d = 1cm = 1× 10–3m


voltage V = 100V


Thickness t = 0.5 = 5 × 10–4m


dielectric constant, k = 5


The capacitance of the capacitor without the dielectric slab is given by


Where,


A= Plate Area


d= separation between the plates,


0 = Permittivity of free space = 8.854 × 10-12 m-3 kg-1 s4 A2




When the dielectric slab is inserted, the capacitance becomes



where, t is the thickness of the slab






Question 56.

Find the capacitances of the capacitors shown in figure. The plate area is A and the separation between the plates is d. Different dielectric slabs in a particular part of the figure are of the same thickness and the entire gap between the plates is filled with the dielectric slabs.




Answer:

a)The capacitors are as shown in the fig



Here, the two parts of the capacitor


are in series


Capacitances C1 and C2 with dielectric constants as K1 and K2



Where,


A= Plate Area


d= separation between the plates,


0 = Permittivity of free space = 8.854 × 10-12 m-3 kg-1 s4 A2


k = dielectric strengthof the material


Here,


and


Since, the distance between the plates is divided into two parts,


hence, separation between the plates becomes =


and



Because they are in series, the equivalent capacitance is


calculated as:



Substituting the values,




Here, the capacitor has three parts. These can be taken in series.


b)



Now, in this case, there are three capacitors connected as shown in fig.


These capacitors are connected in series.


capacitance c is given by –



Where,


A= Plate Area


d= separation between the plates,


0 = Permittivity of free space = 8.854 × 10-12 m-3 kg-1 s4 A2


k = dielectric strengthof the material


Capacitors are as follows –


Since, the entire distance is separated into three parts,




Similarly, the other two capacitors




These three capacitors are connected in series


Thus, the net capacitance is calculated as-






c) Here, the capacitors are connected as shown in fig.



We know, capacitance c is given by-



Where,


A= Plate Area


d= separation between the plates,


0 = Permittivity of free space = 8.854 × 10-12 m-3 kg-1 s4 A2


k = dielectric strengthof the material


Capacitors C1 andC2 is given by-





These two capacitors are connected in parallel, net capacitance


becomes





Question 57.

A capacitor is formed by two square metal-plates of edge a, separated by a distance d. Dielectrics of dielectric constants K1 and K2 are filled in the gap as shown in figure. Find the capacitance.




Answer:

These two capacitors are connected in series.



To find out the capacitance, let us consider a small capacitor of


differential width dx at a distance x from


the left end of the capacitor.


The two capacitive elements of dielectric


constants K1 and K2 are with plate


separations as -


and in series,


respectively as seen from fig.


Also, differential plate areas of the capacitors are adx.


We know, capacitance c is given by-



Where,


A= Plate Area


d= separation between the plates,


0 = Permittivity of free space = 8.854 × 10-12 m-3 kg-1 s4 A2


k = dielectric strengthof the material


Then, looking into the fig, the capacitances of the capacitive elements of the elemental capacitors are given by –


and



We know that equivalent capacitance of capacitors connected in


series is given by the expression –





Now, integrating both sides to get the actual capacitance,







Looking back into the fig.



Substituting in the expression for capacitance C,



Now,






Question 58.

Shows two identical parallel plate capacitors connected to a battery through a switch S. Initially, the switch is closed so that the capacitors are completely charged. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant 3. Find the ratio of the initial total energy stored in the capacitors to the final total energy stored.




Answer:

Initially the switch is closed and the capacitors are fully charged.


When the switch is closed, both capacitors are in parallel as shown in fig,


Hence the total energy stored by the capacitor when switch is closed is –




where C is the capacitance and V is the applied voltage.


When the switch is opened and dielectric is induced, the capacitance is




Given dielectric constant as 3



The total energy stored by the capacitor when switch is closed is –


For capacitor at AB




And EBC is –




Total energyEf)





Now, the ratio of the initial total energy stored in the capacitors to the final total energy stored –



=3/5



Question 59.

A parallel-plate capacitor of plate area A and plate separation d is charged to a potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. Find the work done on the system in the process of inserting the slab.


Answer:


Before inserting slab-


We know, capacitance c is given by-



Where,


A= Plate Area


d= separation between the plates,


0 = Permittivity of free space = 8.854 × 10-12 m-3 kg-1 s4 A2


Energy stored by the capacitor




where C is the capacitance and V is the applied voltage.


After inserting slab capacitance c is given by-



Where,


A= Plate Area


d= separation between the plates,


0 = Permittivity of free space = 8.854 × 10-12 m-3 kg-1 s4 A2


k = dielectric strengthof the material


Also, the final voltage becomes


Energy stored by the capacitor–



where ,


C is the capacitance and V is the applied voltage, k is the dielectric constant of the material


The work done on the system in the process of inserting the slab


= change in energy






Question 60.

A capacitor having a capacitance of 100 μF is charged to a potential difference 50V.

a) What is the magnitude of the charge on each plate?

b) The charging battery is disconnected and a dielectric of dielectric constant 2.5 is inserted. Calculate the new potential difference between the plates.

c) What charge would have produced this potential difference in absence of the dielectric slab.

d) Find the charge induced at a surface of the dielectric slab.


Answer:

Given –


Capacitance, C = 100 μF


Potential difference, V = 50V.


a) We know the magnitude of the charge on each plate is given by


Charge stored on the capacitor, q = c × v


where c is the capacitance and v is the potential difference


⇒ q = 100 μF×50


=5 mC


b)Now, the charging battery is disconnected and a dielectric of dielectric constant 2.5 is inserted.


The new potential difference between the plates will be –





c)


Now, the charge on the capacitance can be calculated as:


Charge, q= Capacitance, C × Potential difference, V


Putting the value in the above formula, we get


Q= 20 × 100 × 10-6 =2 mC


d)The charge induced at a surface of the dielectric slab –



Where, qi is the induced charge, q is the initial charge and k is the dielectric constant of the material inserted.





Question 61.

A spherical capacitor is made of two conducting spherical shells of radii a and b. The space between the shells is filled with a dielectric of dielectric constant K up to a radius c as shown in figure. Calculate the capacitance.




Answer:

Lets re-draw the diagram-


We know, capacitance for a spherical capacitance c is given by-


C =



Where,


C: Capacitance


ri: inner radius


ro: outer radius


k: relative permittivity


∈: permittivity of space


Capacitance between c and a-



Similarly, between b and c



From fig, we can see that the two capacitors are connected in series, hence the net capacitance is given by-







Question 62.

Consider an assembly of three conducting concentric spherical shells of radii a, b and c as shown in figure. Find the capacitance of the assembly between the points A and B.


Answer:

These three metallic hollow spheres form two spherical capacitors, which are connected in series.


Solving them individually, for 1) and 2)


For a spherical capacitor formed by two spheres of radii ro > ri is given by


C =


Where,


C: Capacitance


ri: inner radius


ro: outer radius


k: relative permittivity or dielectric constant


∈: permittivity of space


Similarly,



And




Now, the capacitors are connected in series, net capacitance for series connected capacitors is given by –







Question 63.

Suppose the space between the two inner shells of the previous problem is filled with a dielectric of dielectric constant K. Find the capacitance of the system between A and B.




Answer:

The two capacitors are connected in series, hence the net capacitance is given by



For a spherical capacitor formed by two spheres of radii ro > ri is given by


C =


Where,


C: Capacitance


ri: inner radius


ro: outer radius


k: relative permittivity or dielectric constant


∈: permittivity of space


Similarly,



And







Question 64.

An air-filled parallel-plate capacitor is to be constructed which can store 12 μC of charge when operated at 1200V. What can be the minimum plate area of the capacitor? The dielectric strength of air is 3 × 106 V m–1.


Answer:

Given,


Q = 12μc =


V = 1200V


dielectric strength, b = 3 x 106V/m


Plate Area can be calculated as follows –


The separation between the plates of the capacitor is given by-



where v is the applied voltage and b is the dielectric strength




Now, the capacitance of the capacitor is given by



where Q is the charge stored and V is the voltage applied.




We know, capacitance c is given by-



Where,


A= Plate Area


d= separation between the plates,


0 = Permittivity of free space = 8.854 × 10-12 m-3 kg-1 s4 A2


Thus, the area of the plates is given by –




⇒ A= m2


=0.45 m2



Question 65.

A parallel-plate capacitor with the plate area 100 cm2 and the separation between the plates 1.0 cm is connected across a battery of emf 24 volts. Find the force of attraction between the plates.


Answer:

Given -


Area of the plates of the capacitor, A = 100 cm2 = 10-2 m2


Separation between the plates, d = 1 cm = 10-2 m


Emf of battery, V = 24 V


We know, capacitance c is given by-



Where,


A= Plate Area


d= separation between the plates,


0 = Permittivity of free space = 8.854 × 10-12 m-3 kg-1 s4 A2


Therefore,



Capacitance,


=




Energy stored by the capacitor





Now, force of attraction between the plates,



where


E = energy stored and d is the separation between the plates






Question 66.

Consider the situation shown in figure. The width of each plate is b. The capacitor plates are rigidly clamped in the laboratory and connected to a battery of emf Є. All surfaces are frictionless. Calculate the value of M for which the dielectric slab will stay in equilibrium.




Answer:

Let the battery connected to the capacitor be of potential V.


Let the length of the part of the slab inside the capacitor be x.


b – Width of plates


The capacitor can be considered to be two capacitors which are connected in parallel.



The capacitances of the two capacitors in parallel is given by –




C1 is the part of the capacitor having the dielectric inserted in it and C2 is the capacitance of the part of the capacitor without dielectric.


As, C1 and C2 are in parallel therefore, the net capacitance is given by




This dielectric slab is attracted by the electric field of the capacitor and applies a force


The direction of force is in left direction.



Let assume that electric force of magnitude F pulls the slab toward left direction.



Let there be an differential displacement dx towards the left direction by the force F.



The work done by the force




Let V1 and V2 be the potential of the battery connected to the left capacitor and that of the battery connected with the right capacitor


With increase in the displacement of slab, the capacitance will increase, hence the energy stored in the capacitor will also increase.


In order to maintain constant voltage, the battery will supply extra charge, and gets damage .


Therefore the battery will do work.


Now,


Work done by the battery


= Energy change of capacitor + work done by the force F on the capacitor



1)



Let’s take the differential charge dq is supplied by the battery, and the change in the capacitor be dC


We know that energy in capacitor dWB




we know q = cv


2)



And force F is given by,




From 1) and 2)






In order to keep the dielectric slab in equilibrium, the electrostatic force acting on it must be balanced by the weight of the block attached.


Therefore,





Question 67.

Figure shows two parallel plate capacitors with fixed plates and connected to two batteries. The separation between the plates is the same for the two capacitors. The plates are rectangular in shape with width b and lengths ℓ1 and ℓ2. The left half of the dielectric slab has a dielectric constant K1 and the right half K2. Neglecting any friction, find the ratio of the emf of the left battery to that of the right battery for which the dielectric slab may remain in equilibrium.




Answer:

Let V1,V2 be the potential of the battery connected to the left capacitor and that of the battery connected to the right capacitor



Considering the left capacitor -


Let the length of the part of the slab inside the capacitor be x.


The left capacitor can be considered to be two capacitors in parallel.


Let the battery connected to the capacitor be of potential V.


Let the length of the part of the slab inside the capacitor be x.


b – Width of plates


The capacitances of the two capacitors in parallel is given by –




C1 is the part of the capacitor having the dielectric inserted in it and C2 is the capacitance of the part of the capacitor without dielectric.


As, C1 and C2 are in parallel therefore, the net capacitance is given by





Therefore, the potential energy stored in the left capacitor will be



1)


This dielectric slab is attracted by the electric field of the capacitor and applies a force.



Let assume that electric force of magnitude F pulls the slab toward left direction.



Let there be an differential displacement dx towards the left direction by the force F.



The work done by the force




Let V1 and V2 be the potential of the battery connected to the left capacitor and that of the battery connected with the right capacitor


With increase in the displacement of slab, the capacitance will increase, hence the energy stored in the capacitor will also increase.


Let us consider a small displacement dx of the slab towards the inward direction.


In order to maintain constant voltage, the battery will supply extra charge, and gets damage .


Therefore the battery will do work.


Now,


Work done by the battery


= Energy change of capacitor + work done by the force F on the capacitor



1)


Let’s take the differential charge dq is supplied by the battery, and the change in the capacitor be dC


We know that energy in capacitor dWB




we know q = cv


2)



And force F is given by,




From 1) and 2)






Solving for voltages V1 and V2 -




Similarly, for the right side the voltage of the battery is given by-



Now, the ratio of the voltages is given by-




Thus, the ratio of the emfs of the left battery to the right battery is given by -




Question 68.

Consider the situation shown in figure.

The plates of the capacitor have plate area A and are clamped in the laboratory. The dielectric slab is released from rest with a length a inside the capacitor. Neglecting any effect of friction or gravity, show that the slab will execute periodic motion and find its time period.




Answer:

Given



area of the plates of the capacitors = A.



a = length of the dielecric slab is inside the capacitor.



Therefore, the area of the plate covered with dielectric is =



The capacitance of the portion with dielectric is given by



The capacitance of the portion without dielectric is given by



The two parts can be considered to be in parallel.



Therefore, the net capacitance is given by-





Let us consider a small displacement da of the slab towards the inward direction.


With increase in the displacement of slab, the capacitance will increase, hence the energy stored in the capacitor will also increase.


In order to maintain constant voltage, the battery will supply extra charge, and gets damage .


Therefore the battery will do work.


Now,


Work done by the battery


= Energy change of capacitor + work done by the force F on the capacitor



1)


Let’s take the differential charge dq is supplied by the battery, and the change in the capacitor be dC


We know that energy in capacitor dWB




we know q = cv


2)



And force F is given by,




From 1) and 2)






We know


2)


where m is the mass of the object


a is the acceleration


From 1) and2)-


the acceleration of the dielectric a0 is given by =



where, m is the mass.


As, the force is in inward direction, it tends to make the dielectric to completely fill the space inside the capacitors.


As, the dielectric tends to completely fills the space inside the capacitor, at this instant its velocity is not zero.


After that the dielectric slab tends to move outside the capacitor.


As the slab tends to move out, the direction of force reverses.


Hence, the dielectric slab will maintain periodic motion.


Now, the time required for moving a distance l-a) can be-





=


For completing cycle, the time taken will be four times the time taken for covering distance l-a)



hence,


T=4t


=4×


=8