ROUTERA


Electric Field And Potential

Class 12th Concepts Of Physics Part 2 HC Verma Solution



Short Answer
Question 1.

The charge on a proton is +1.6 × 10–19 C and that on an electron is –1.6 × 10–19 C. Does it mean that the electron has a charge 3.2 × 10–19C less than the charge of a proton?


Answer:


No, it does not mean that the electron has a charge 3.2 × 10–19C less than the charge of a proton. Electrons and Protons have same amount of charge just the nature of charge is different. Electron has negative charge and Proton has positive charge. If we keep an electron and a proton at a distance apart then attractive force will be observed as their nature of charge is different.
Hence, magnitude of charge of both electron and a proton is
1.6 × 10–19 C but nature is opposite.



Question 2.

Is there any lower limit to the electric force between two particles placed at a separation of 1 cm?


Answer:


Yes, there is a lower limit to the electric force between two particles.
We know that electric force is given as:
This is Coulomb’s Law. Here, k is a constant .
k = = 9× 109 Nm2C-2, q1 and q2 are the charges of two particles and r is the distance between two charges.
The smallest possible charge would be that of an electron.
r=1 cm = 0.01 m
Thus,

Hence, lower limit to the electric force between two particles placed at a separation of 1 cm would be 2.3× 10-24 N.



Question 3.

Consider two particles A and B having equal charges and placed at some distance. The particle A is slightly displaced towards B. Does the force on B increase as soon as the particle A is displaced? Does the force on the particle An increase as soon as it is displaced?


Answer:


Yes, the force on particle B as well as particle A will increase when particle is A is displaced towards particle B.
By Coulomb’s Law, Electric force is inversely proportional to the square of the distance between the two charges.
F α 1/r2
Thus, when A is displaced towards B, the distance between them decreases and hence the force on both the particles will increase.



Question 4.

Can a gravitational field be added vectorially to an electric field to get a total field?


Answer:


No, gravitational field cannot be added vectorially to an electric field to get a total field. This is because Electric field comprises of influence due to electric charges whereas gravitational field comprises of influence due to masses of the bodies and thus, electric field and gravitational field have different dimensions. We can obtain net Force by adding Gravitational force and electric force.
Hence, gravitational field cannot be added to electric field vectorially.



Question 5.

Why does a phonograph-record attract dust particles just after it is cleaned?


Answer:


When a phonograph record is cleaned, electric charges are produced on the record due to induction. This happens as we rub a cloth on the record, friction causes the induction of charges on the record. This phenomenon is similar to a glass rod being rubbed by a cloth. Now, these deposited charges on the record attract dust particles having neutral charge.
Hence, a phonograph-record attract dust particles just after it is cleaned.



Question 6.

Does the force on a charge due to another charge depend on the charges present nearby?


Answer:


No, electric force between two charges or on any one of the two charges does not depends on the charges present nearby.
According to Coulomb’s Law,
Electric force is given as:
Here, k is a constant and k = = 9× 109 Nm2C-2, q1 and q2 are the two charges and r is the distance between two charges. As we see, the force on one charge depends only on the magnitude of the second charge and vice versa.
Third charge won’t affect the force between first two charges.
The only thing that would affect the force on the charge is the net force due to superposition of all the individual forces.


Hence, the force on a charge due to another charge does not depend on the charges present nearby.



Question 7.

In some old texts it is mentioned that 4π lines of force originate from each unit positive charge. Comment on the statement in view of the fact that 4π is not an integer.


Answer:


Here 4π is considered as a solid angle. Meaning the lines of force can subtend radially outward as shown in the figure below:

It does not mean that 4π lines will come out of a point charge, it means lines of force will move radially outwards from a positive charge having radial angle of 4π.



Question 8.

Can two equipotential surfaces cut each other?


Answer:


No, two equipotential surfaces cannot cut each other. When two equipotential surfaces intersect at a point, the potential at that point will have two values which is not possible. Also, electric field is perpendicular to the equipotential surface, when two surfaces intersect there will be two directions of the electric field ( one of each equipotential surface) which is not possible either.
Hence, two equipotential surfaces cannot cut each other.



Question 9.

If a charge is placed at rest in an electric field, will its path be along a line of force? Discuss the situation when the lines of force are straight and when they are cured.


Answer:


Yes, when a charge is placed at rest in an electric field its path will be tangential the line of force. When the lines of forces are straight, the path of the charge would be in a straight line. When the lines of force are curved, the path of the charge would be tangential to the curve.



Question 10.

Consider the situation shown in figure. What are the signs of q1 and q2? If the lines are drawn in proportion to the charge, what is the ratio q1/q2?





Answer:


Lines of force are moving outwards from q2, hence q2 is positively charged: +q2. Lines of force are going into q1, hence q1 is negatively charged : -q1.
From the diagram, 18 lines of electric field come out of q2 and 6 lines of electric field get into q1.
Ratio of
Hence, If the lines are drawn in proportion to the charge then q1:q2= 1:3



Question 11.

A point charge is taken from a point A to a point B in an electric field. Does the work done by the electric field depend on the path of the charge?


Answer:


No, the work done by the electric field does not depend on the path of the charge.
Work done is given as: W = Potential difference × charge.
= (VA – VB) × q
Here, VA and VB is the potential at Points A and B respectively. And q is the charge.
Thus, the position of the charge is important as potential difference will vary with position and so the work done will change.
Hence, when a point is taken from A to a point B the work done on the charge due to electric field will not depend on the path.



Question 12.

It is said that the separation between the two charges forming an electric dipole should be small. Small compared to what?


Answer:


The distance between the two charges forming the dipole must be small compared to the distance between the point of influence and the center of the dipole.



Question 13.

The number of electrons in an insulator is of the same order as the number of electrons in a conductor. What is then the basic difference between a conductor and an insulator?


Answer:


The basic difference between a conductor and an insulator is that conductor has many free electrons in its outer shell and an insulator has no free electrons. Free electrons are responsible for conduction of electricity in a material as they are not bound to the atom and are free to move. Whereas in an insulator electrons are tightly bound to the atom, thus no free electrons.



Question 14.

When a charged comb is brought near a small piece of paper, it attracts the piece. Does the paper become charged when the comb is brought near it?


Answer:


When a charged comb is brought near a small piece of paper, it attracts the piece because of induction. Induction is the process of redistribution of the charges on one body due to the presence of a charged body. Similarly, the charges are distributed on the paper from the comb. Thus when a charged comb is brought near a piece, the charges at the pointing end of the comb attracts the opposite charge on the piece of the paper which were introduced due to induction.
Whole paper does not become charged, just the area near the charged comb contains induced electrons.

Hence When a charged comb is brought near a small piece of paper, it attracts the piece.




Objective I
Question 1.

Figure shows some of the electric field lines corresponding to an electric field. The figure suggests that



A. EA > EB > EC

B. EA = EB = EC

C. EA = EC > EB

D. EA = EC < EB


Answer:

At any given point in the electric field region, the electric flux (i.e. the number of field lines per unit area) determines the intensity of the field. The points in the region where the flux is more are associated with strong electric fields.


For a given area, field lines are dense and equal in number at point A and C.


Therefore, EA = EC


And at point B, the field lines are far away which indicates lesser intensity of electric field at point B than at point A and C.


Question 2.

When the separation between two charges is increased, the electric potential energy of the charges
A. increases

B. decreases

C. remains the same

D. may increase or decrease


Answer:

Electric potential energy between two charges Q and Q’ kept at a distance R is given by:



If R is increased, then energy may decrease or increase depending on the charges involved.


∴ The potential energy may increase or decrease.


Question 3.

If a positive charge is shifted from low-potential region to a high-potential region, the electric potential energy
A. increases

B. decreases

C. remains the same

D. may increase or decrease.


Answer:

Electric potential energy of a system of point charges is defined as the work required in assembling this system of charges by bringing them close together, as in the system from infinite distance.


As we know,


E=-V


Electric field flows from high potential to low potential and a positive charge moves in the direction of electric field. In this process, it loses its energy.


But when a positive charge is moved from low potential to high potential i.e. in the direction opposite to the electric field, its energy increases as some work needs to be done to the particle. This work done gets stored as energy in the particle.


Question 4.

Two equal positive charges are kept at points A and B. The electric potential at the points between A and B (excluding these points) is studied while moving from A to B. The potential
A. continuously increases

B. continuously decreases

C. increases then decreases

D. decrease then increases


Answer:

Potential due to a point charge at a distance r is given by:




Therefore, at a point very nearby to A and B, r → 0 ∴ V → ∞


i.e.


At a very nearby point right to A and a very nearby point left to B, potential will tend to ∞ (maximum possible value) as separation will tend to zero. That means at a point somewhere in the middle of path AB, the potential will acquire its minimum value.


Therefore, while moving from A to B, potential will first decrease and then increase.


Question 5.

The electric field at the origin is along the positive x-axis. A small circle is drawn with the center at the origin cutting the axes at points A, B, C and D having coordinates (α, 0), (0, α), (–α, 0), (0, –α) respectively. Out of the points on the periphery of the circle, the potential is minimum at



A. A

B. B

C. C

D. D


Answer:

As the electric field is in +X direction, that is field lines will flow from point C to point A. And as we already know that field lines flow from high potential to low potential, that means point A needs to be at a low potential region and point C needs to be at a high potential region.


Points B and D are equipotential.


The order of potentials is:


VC > VB =VD > VA


Therefore, potential is minimum at A.


Question 6.

If a body is charged by rubbing it, its weight
A. remains precisely constant

B. increases slightly

C. decreases slightly

D. may increase slightly or may decrease slightly.


Answer:

On charging a body by rubbing it, there are two possibilities to happen: Either the body acquires a positive charge or a negative charge.


As q=ne (where q is the amount of charge acquired on a body, n is the number of electrons transferred from or on the body and e is the charge on one electron.)


In case the body acquires a positive charge, there is loss of electrons from that body, therefore, its mass decreases.


In case the body acquires a negative charge, there is gain of electrons from that body, therefore, its mass increases.


So, on charging a body, its mass may increase or may decrease slightly.


Question 7.

An electric dipole is placed in a uniform electric field. The net electric force on the dipole
A. is always zero

B. depends on the orientation of the dipole

C. can never be zero

D. depends on the strength of the dipole


Answer:

An electric dipole consists of equal and opposite charges placed at some distance.



The dipole moment of this configuration is given as p=q(2a)


When the dipole is kept in an external electric field E, both its charges experience some force. Let the force on the positive charge(q) and negative charge(-q) be FP and FN respectively.


Therefore, FP =qE and FN =-qE.


I.e. FP =- FN


that is force on the dipole is equal in magnitude but opposite in direction irrespective of its orientation.


Therefore, force on an electric dipole placed in an electric field is always zero.


Question 8.

Consider the situation of figure. The work done in taking a point charge from P to A is WA, from P to B is WB and from P to C is WC.



A. WA < WB < WC

B. WA > WB > WC

C. WA = WB = WC

D.None of these


Answer:

Since electric field is a conservative field and therefore electric force is also a conservative force. Therefore, work done will not depend in the path taken.


As for a conservative force field, work done on a charge does not depend on the path taken by the charge but depends only on the initial and final points. Therefore, in this situation,


i.e. WA = WB = WC


∴ The work done in taking a point charge from P to all the points A, B and C is the same.


Question 9.

A point charge q is rotated along a circle in the electric field generated by another point charge Q. The work done by the electric field on the rotating charge in one complete revolution is
A. zero

B. positive

C. negative

D. zero if the charge Q is at the center and nonzero otherwise.


Answer:

If a force F applies on a body and it allows it to move an infinitely small distance d, then work done is given by dW=F.ds and for a path, integrate it putting lower limits (starting position) and upper limits (ending position).


Here, in the given question,


since a complete rotation is made i.e. net displacement is zero. i.e.



i.e. The work done by the electric field on the rotating charge in one complete revolution is zero.



Objective Ii
Question 1.

Mark out the correct options.
A. The total charge of the universe is constant.

B. The total positive charge of the universe is constant.

C. The total negative charge of the universe is constant.

D. The total number of charged particles in the universe is constant.


Answer:

Option (a) is correct because total charge of the universe is constant. It just gets transferred from one particle to another.


Option (b) is incorrect as the total positive charge of the universe is not constant. It is because the positive charges get converted into negative charges and vice-versa.


Option (c) is incorrect as the total negative charge of the universe is not constant. It is because the negative charges get converted into positive charges and vice-versa.


Option (d) is incorrect as in universe, the total number of charged particles is not a constant as a pair of positive and negative charges appear at the time of pair production and a pair of positive and negative charge combine to form a neutral particle at the time of pair annihilation.


Question 2.

A point charge is brought in an electric field. The electric field at a nearby point.
A. will increase if the charge is positive

B. will decrease if the charge is negative

C. may increase if the charge is positive.

D. may decrease if the charge is negative.


Answer:

Option (c) is correct:


In an electric field, if a positive charge is placed, then the charge itself will generate some electric field on its own diverging from the point.


At a nearby point Q (towards the direction of external field), the field will increase as the field lines from the charge will also add to external field lines.


At a nearby point P (away from the direction of external field), the field will decrease as the field lines from the charge will cancel some of the external field lines.



Therefore, at a nearby point, field can either increase or decrease.


Option (d) is correct:


In an electric field, if a negative charge is placed, then the charge itself will generate some electric field on its own converging to the point.


At a nearby point Q (towards the direction of external field), the field will decrease as the field lines from the charge will cancel some of the external field lines.


At a nearby point P (away from the direction of external field), the field will increase as the field lines from the charge will add on to the external field lines.



Therefore, at a nearby point, field can either increase or decrease.


Question 3.

The electric field and the electric potential at a point are E and V respectively.
A. If E = 0, V must be zero.

B. If V = 0, E must be zero.

C. If E ≠ 0, V cannot be zero

D. If V ≠ 0, E cannot be zero.


Answer:

We know that electric field is negative of the space rate of change of electric potential in a region. i.e.



Option (a) is incorrect as for E =0, V=0 is not the only condition. As we can see from the above equation, electric field is the negative of vector derivative of V that means


if V=constant, then also, E=0.


(As


Option (b) is incorrect as:


If at a point, the electric potential is zero, it doesn’t imply electric field to be also zero.


Take an example of a dipole where somewhere between the two charges on its axial line, there exists a point where potential is zero but electric field is always non zero and directs in the direction from positive charge to negative charge.


Option (c) is incorrect as if E≠0, it doesn’t mandate the potential to be non-zero. We can take again the example of a dipole where somewhere at a point on its axial line, there exists a point where E≠0 but V=0. So the statement “If E ≠ 0, V cannot be zero” is incorrect.


Option (d) is incorrect as:


If V ≠ 0 (let’s say V= non-zero constant)


Then E=0


Which is contradicted in the given statement.


Thus, none of the options is correct.


Question 4.

The electric potential decreases uniformly from 120V to 80V as one moves on the x-axis from x = –1 cm to x = +1 cm. The electric field at the origin
A. must be equal to 20 V cm–1

B. may be equal to 20 V cm–1

C. may be greater than 20 V cm–1

D. may be less than 20 V cm–1.


Answer:



Here also, and


Upon solving, we get but it will also depend on the angle that equipotential surface makes with the X axis.


Therefore, the value of electric field may greater than 20V/cm (if the potential decreases uniformly) or equal to 20 V/cm (when equipotential surfaces are at right angles from X axis.)


Question 5.

Which of the following quantities do not depend on the choice of zero potential or zero potential energy?
A. Potential at a point

B. Potential difference between two points

C. Potential energy of a two-charge system

D. Change in potential energy of a two-charge system.


Answer:

Option (a) is incorrect because potential at a point depends on our choice of zero potential. Normally infinity is taken as reference point where potential is assumed to be zero.


If choice of zero potential is changed, then the potential at a point will also change.


Option (b) is correct as difference in potential between the two points doesn’t depend on the choice of zero potential.


Option (c) is incorrect as the potential energy of the two-charge system also depends on the choice of zero potential in order to calculate the potential at the point where the two charges are kept.


Option (d) is correct as the change in potential energy of the two-charge system is independent of our choice of the zero-potential point.


Thus, we conclude that potential difference between two points and change in potential energy of a two-charge system don’t depend on the choice of zero potential point.


Question 6.

An electric dipole is placed in an electric field generated by a point charge.
A. The net electric force on the dipole must be zero.

B. The net electric force on the dipole may be zero.

C. The torque on the dipole due to the field must be zero.

D. The torque on the dipole due to the field may be zero.


Answer:

The field due a point charge is always radial (radially outwards for positive charge and radially inwards for negative charge). A dipole placed in that electric field will never feel equal and opposite forces due to the radial nature of the field. Therefore, option (a) and (b) are incorrect.


Torque on the dipole may be zero in the case when the dipole is placed along the electric field (as in that only case, and will be parallel or antiparallel, thus torque will be zero.)


Question 7.

A proton and an electron are placed in a uniform electric field.
A. The electric forces acting on them will be equal.

B. The magnitudes of the forces will be equal.

C. Their accelerations will be equal.

D. The magnitudes of their accelerations will be equal


Answer:


Let the force on the electron and proton is Fe (F) and Fp (F’) respectively.


Then,



Option (a) is incorrect as the forces are equal only in magnitude but not in direction.


Option(b) is correct as magnitude of the forces are equal.


Option (c) is incorrect as the magnitude of accelerations for both particles is not equal and also the direction is also not the same.


Option (d) is incorrect as the magnitude of acceleration of both will be different as the masses of both are different.


Question 8.

The electric field in a region is directed outward and is proportional to the distance r from the origin. Taking the electric potential at the origin to be zero,
A. it is uniform in the region

B. it is proportional to r

C. it is proportional to r2

D. it increases as one goes away from the origin


Answer:

The field around the origin will be like:



Where the intensity of the field will increase as we will move away from the charge placed at origin (as field is proportional to the distance r from the origin.)


Option (a) is incorrect as the field is not uniform in the region (can be perceived by the given situation and the diagram)


Field is given as:



(where k is some positive constant)


Now,





Option (b) is incorrect since V is not proportional to r.


Option (c) is correct since V is proportional to r.


Option (d) is incorrect. From the expression for V, we can say that the potential decreases as we move away. Alternatively, the field lines always flow from high potential to low potential. Therefore, we can say that the potential decreases with increase in r.



Exercises
Question 1.

Find the dimensional formula of ϵ0.


Answer:

We know that,


Electrostatic force,


Where


Єo is the permittivity of vacuum


q1, q2 is the magnitude of the charges on the charge particles


r is the distance between the two charge particles.


Taking the dimensions,



Note that Charge = Current × Time. Hence, its dimensional formula is


Note that 4π is a dimensionless constant. Note that . Hence, . Also, r is just distance or length. Thus,



Rearranging, we get




Question 2.

A charge of 1.0 C is placed at the top of your college building and another equal charge at the top of your house. Take the separation between the two charges to be 2.0 km. Find the force exerted by the charges on each other. How many times of your weight is this force?


Answer:

Given:



Distance between the charges,


Formula used:


By Coulomb’s law, the electric force is given by:



Where ϵ0 is the permittivity of free space


q1 and q2 are the magnitude of charges


r is the distance of separation between the charges


By Coulomb’s law,





Formula used:


The weight of an object is given by



where g is the acceleration due to gravity


and m is the mass of the object.


Let the mass of my body, m be 70 kg.



Thus, dividing the Electrostatic force and the body weight, we get



The electric force between the two charges is 3.3 times my weight.



Question 3.

At what separation should two equal charges, 1.0C each, be placed so that the force between them equals the weight of a 50 kg person?


Answer:

Given:


Mass of the person


Weight of the person,


Charges:


Formula used:


By Coulomb’s law, the electric force is given by:



Where ϵ0 is the permittivity of free space


q1 and q2 are the magnitude of charges


r is the distance of separation between the charges


__________________________________________________


The magnitude of electric force is given by:



Rearranging, we get



Note that Substituting all the values, we get





Question 4.

Two equal charges are placed at a separation of 1.0 m. What should be the magnitude of the charges so that the force between them equals the weight of a 50 kg person?


Answer:

Given:


Mass of the person


Weight of the person,


Let the magnitude of charge on both be q.


Here,


Distance of separation,


________________________________________________


Formula used:


By Coulomb’s law, the electric force is given by:



Where ϵ0 is the permittivity of free space


k is the electrostatic constant


q1 and q2 are the magnitude of charges


r is the distance of separation between the charges


__________________________________________________




Rearranging, we get



Substituting the corresponding values, we get





Question 5.

Find the electric force between two protons separated by a distance of 1 fermi (1 fermi = 10–15 m). The protons are a nucleus remain at a separation of this order.


Answer:

Given:


Charge on each proton, q1=q2=1.6×10-19 C


Distance of separation,


________________________________________________


Formula used:


By Coulomb’s law, the electric force is given by:



Where ϵ0 is the permittivity of free space


k is the electrostatic constant


q1 and q2 are the magnitude of charges


r is the distance of separation between the charges


__________________________________________________


The magnitude of electric force is given by:






Question 6.

Two charges 2.0 × 10–6 C and 1.0 × 10–6 C are placed at a separation of 10 cm. Where should a third charge be placed such that it experiences no net force due to these charges?


Answer:

Given:


Charges:


Distance between the charge is 10 cm


Let the third charge be q. There are three possible scenarios:



As we can see from the picture, the only possible location for q where net force on q is zero is between the two charges. Let it be x m away from q1, x > 0.



________________________________________________


Formula used:


By Coulomb’s law, the electric force is given by:



Where ϵ0 is the permittivity of free space


k is the electrostatic constant


q1 and q2 are the magnitude of charges


r is the distance of separation between the charges


__________________________________________________


Force on q due to q1 is given by:



Force on q due to q2 is given by:



Now,








As x > 0. The correct option is 0.041.


Distance from charge q2 = 0.1 – 0.041 = 0.059 m


Hence, the charge q should be placed 5.9 cm from q2.



Question 7.

Suppose the second charge in the previous problem is –1.0 × 10–6 C. Locate the position where a third charge will not experience a net force.


Answer:

Given:


Charges: q1=-1.0×10-6 C, q2=2.0×10-6 C



Let the third charge q be at a distance x cm from q2. As the forces must cancel, case II is not possible. Moreover, for forces to cancel, q must be near the smaller charge. Hence, only Case III is possible.



________________________________________________


Formula used:


By Coulomb’s law, the electric force is given by:



Where ϵ0 is the permittivity of free space


k is the electrostatic constant


q1 and q2 are the magnitude of charges


r is the distance of separation between the charges


__________________________________________________


Force on q due to q1 is given by:



Force on q due to q2 is given by:



Now,








For Case III, x must be greater than 10. Hence, x = 34.14 cm


Thus, q should be placed 34.1 cm from the larger charge on the side of the smaller charge.



Question 8.

Two charged particles are placed at a distance 1.0 cm apart. What is the minimum possible magnitude of the electric force acting on each charge?


Answer:

Given: missing


For the electric force to be minimum, the magnitude of charge on the particles must be minimum. As both the particles are charged, they must have non-zero minimum charge.


Hence, they both must carry charge having magnitude of fundamental unit of charge i.e. 1.6 × 1019 C.


Magnitude of charges,


Distance between the charged particles,


________________________________________________


Formula used:


By Coulomb’s law, the electric force is given by:



Where ϵ0 is the permittivity of free space


k is the electrostatic constant


q1 and q2 are the magnitude of charges


r is the distance of separation between the charges


__________________________________________________


The magnitude of electric force is:







Question 9.

Estimate the number of electrons in 100g of water. How much is the total negative charge on these electrons?


Answer:

Given:


Mass of water ,


Molar mass of water


Moles of water in 100g of water =


Number of electrons in 1 molecule of water = 2 times one electron each in hydrogen atom + 8 electrons in oxygen atom = 10 electrons


__________________________________________________


Formulas used:


The relation between molar mass M, mass of sample m and number of moles of sample n is given by:



1 mol of “something” contains NA number of “something”.


__________________________________________________


Number of electrons in 1 mol of electrons =


Number of electrons in 1 g of water


= 10 electrons × Na ×


Number of electrons in 100 g of water





Negative charge on one electron =


Total negative charge in 100g of water



Question 10.

Suppose all the electrons of 100g water are lumped together to form a negatively charged particle and all the nuclei are lumped together to form a positively charged particle. If these two particles are placed 10.0 cm away from each other, find the force of attraction between them. Compare it with your weight.


Answer:

Given:


Mass of water ,


Molar mass of water


Moles of water in 100g of water =


Number of electrons in 1 molecule of water = 2 times one electron each in hydrogen atom + 8 electrons in oxygen atom = 10 electrons


__________________________________________________


Formulas used:


The relation between molar mass M, mass of sample m and number of moles of sample n is given by:



1 mol of “something” contains NA number of “something”.


__________________________________________________


Number of electrons in 1 mol of electrons =


Number of electrons in 100 g of water





Negative charge on one electron =


Total negative charge


We have,


Charges:


Distance of separation,


________________________________________________


Formula used:


By Coulomb’s law, the electric force is given by:



Where ϵ0 is the permittivity of free space


k is the electrostatic constant


q1 and q2 are the magnitude of charges


r is the distance of separation between the charges


__________________________________________________


The magnitude of electric force is:





Let my mass m be 70 kg.




The electric force between the two charges is about 1022 times my weight!



Question 11.

Consider a gold nucleus to be a sphere of radius 6.9 fermi in which protons and neutrons are distributed. Find the force of repulsion between two protons situated at largest separation. Why do these protons not fly apart under this repulsion?


Answer:

Given:


Largest Distance of Separation = Diameter of the nucleus ,


Charge on each proton ,


________________________________________________


Formula used:


By Coulomb’s law, the electric force is given by:



Where ϵ0 is the permittivity of free space


k is the electrostatic constant


q1 and q2 are the magnitude of charges


r is the distance of separation between the charges


The electric force is repulsive and is given by:





The protons do not fly apart because there is also strong nuclear force which is attractive and balances the repulsive electric force.



Question 12.

Two insulating small spheres are rubbed against each other and placed 1 cm apart. If they attract each other with a force of 0.1N, how many electrons were transferred from one sphere to the other during rubbing?


Answer:

Given,


Electric force between the two spheres,


Distance between the two spheres,


We know that, magnitude of charge on electron,


Let n electrons be transferred from one sphere to another. Then, charge on one sphere is +ne and the other is -ne.



________________________________________________


Formula used:


By Coulomb’s law, the electric force is given by:



Where ϵ0 is the permittivity of free space


k is the electrostatic constant


q1 and q2 are the magnitude of charges


r is the distance of separation between the charges


__________________________________________________


The magnitude of electric force is given by:







Hence, about electrons are transferred.



Question 13.

NaCl molecule is bound due to the electric force between the sodium and the chlorine ions when one electron of sodium is transferred to chlorine. Taking the separation between the ions to be 2.75 × 10–8 cm, find the force of attraction between them. State the assumptions (if any) that you have made.


Answer:

Given:


Charge on Na+ ion =


Charge on Cl- ion =


Distance between the ions,


________________________________________________


Formula used:


By Coulomb’s law, the electric force is given by:



Where ϵ0 is the permittivity of free space


k is the electrostatic constant


q1 and q2 are the magnitude of charges


r is the distance of separation between the charges


__________________________________________________


We assume that the distance between the transferred electron and the sodium nucleus is the distance between the two ions.






Question 14.

Find the ratio of the electric and gravitational forces between two protons.


Answer:

Given:


Mass of proton,


Charge on proton,


________________________________________________


Formula used:


By Coulomb’s law, the electric force is given by:



Where ϵ0 is the permittivity of free space


k is the electrostatic constant


q1 and q2 are the magnitude of charges


r is the distance of separation between the charges


__________________________________________________


________________________________________________


Formula used:


By Newton’s law of gravitation, the gravitational force is given by:



Where G is the Gravitational constant


m1 and m2 are the magnitude of charges


r is the distance of separation between the masses


__________________________________________________


The electric force is given by:



The gravitational force is given by:



Dividing (2) by (1), we get





Hence the ratio of electric force to gravitational force between two protons is about 1.23 × 1036.



Question 15.

Suppose an attractive nuclear force acts between two protons which may be written as F = CE–kx/r2.

(a) Write down the dimensional formulae and appropriate SI units of C and κ.

(b) Suppose that κ = 1 fermi–1 and that the repulsive electric force between the protons is just balanced by the attractive nuclear force when the separation is 5 fermi. Find the value of C.


Answer:

Given:


The attractive nuclear force between two protons is,


(a)The power of e must be dimensionless.


Thus,




Hence, SI unit of k is m.


Now,





Let us replace the formula with SI units



Hence, SI unit of C is Nm2


(b)


Given,


Separation between the protons, r = 5fermi = 5 × 10-15 m.


We know that the charge on a proton is q = 1.6 × 10-19C.


Now,


_________________________________________________


Formula used:


By Coulomb’s law, the electric force is given by:



Where ϵ0 is the permittivity of free space


k is the electrostatic constant


q1 and q2 are the magnitude of charges


r is the distance of separation between the charges


__________________________________________________


The electric force between the two protons is given by:



Also, the nuclear force between the protons is given by:



These two forces balance each other.


Fe=Fn


kq2=Ce-kr






Question 16.

Three equal charges, 2.0 × 10–6 C each, are held fixed at the three corners of an equilateral triangle of side 5 cm. Find the Coulomb force experienced by one of the charges due to the rest two.


Answer:

Given:


Let the force due to charge at corner B, C be FB, FC.


Charge at each corner, .


Note that the length of side of the triangle is.


________________________________________________


Formula used:


By Coulomb’s law, the electric force is given by:



Where ϵ0 is the permittivity of free space


k is the electrostatic constant


q1 and q2 are the magnitude of charges


r is the distance of separation between the charges


__________________________________________________




Putting the values in the above formula, we get



We can see from the diagram that FCand FBcab be resolved into theirs x and y components.






The y’s are in the same direction and add up but the x’s are in the opposite direction, hence they cancel out.


Hence,





Question 17.

Four equal charges 2.0 × 10–6 C each are fixed at the four corners of a square of side 5 cm. Find the Coulomb force experienced by one of the charges due to the rest three.


Answer:

Given:


Let the force due to charge at corner B, C and D be FB, FC,FD.


Charge at each corner, .


The side of square is.


By Pythagoras theorem,


The length of diagonal .



________________________________________________


Formula used:


By Coulomb’s law, the electric force is given by:



Where ϵ0 is the permittivity of free space


k is the electrostatic constant


q1 and q2 are the magnitude of charges


r is the distance of separation between the charges


__________________________________________________





Resolving FC , we get






The net force is:




We know that



Hence, the magnitude of net force is




Question 18.

A hydrogen atom contains one proton and one electron. It may be assumed that the electron revolves in a circle of radius 0.53 angstrom (1 angstrom = 10–19 m and is abbreviated as ) with the proton at the center. The hydrogen atom is said to be in the ground sate in this case. Find the magnitude of the electric force between the proton and the electron of a hydrogen atom in its ground state.


Answer:

Given:


Magnitude of charge on both proton and electron,


The radius of the circular orbit,



The radius of the circular orbit is the distance between the electron and the proton.


___________________________________________________


Formula used:


By Coulomb’s law, the electric force is given by:



Where ϵ0 is the permittivity of free space


q1 and q2 are the magnitude of charges


r is the distance of separation between the charges


(Here, )


___________________________________________________


The magnitude of electric force is given by:



Putting the values in the above formula, we get




Question 19.

Find the speed of the electron in the ground state of a hydrogen atom. The description of ground state is given in the previous problem.


Answer:

Given:


Magnitude of charge on both proton and electron,


The radius of the circular orbit,


Mass of electron ,



The radius of the circular orbit is the distance between the electron and the proton.


___________________________________________________


Formula used:


By Coulomb’s law, the electric force is given by:



Where ϵ0 is the permittivity of free space


q1 and q2 are the magnitude of charges


r is the distance of separation between the charges


(Here, )


___________________________________________________


The magnitude of electric force is given by:




___________________________________________________


Formula used:


Centripetal force is given by



Where m is the mass if the object, v is the speed of the object and r is the radius of the circular path.


___________________________________________________


Let Fc be the centripetal force on the electron.



Now, the electric force acts as centripetal force.


Hence,



Substituting (1) and rearranging, we get



Putting the values in the above formula, we get




Question 20.

Ten positively charged particles are kept fixed on the x-axis at points x = 10 cm, 20 cm 30 cm, …., 100 cm. The first particle has a charge 1.0 × 10–8 C, the second 8 × 10–8 C, the third 27 × 10–8 C and so on. The tenth particle has a charge 1000 × 10–8 C. Find the magnitude of the electric force acting on a 1C charge placed at the origin.


Answer:

Given:


Charge on the ith particle ,


Distance of ith particle from origin,


________________________________________________


Formula used:


By Coulomb’s law, the electric force is given by:



Where ϵ0 is the permittivity of free space


q1 and q2 are the magnitude of charges


r is the distance of separation between the charges


__________________________________________________


Magnitude of force on the 1C charge due to ithparticle is given by:


(Here, q1 =, and )



Substituting the values, we get



All the forces are in negative-x direction. Hence, we can simply add them up to get the magnitude of net force.






Question 21.

Two charged particles having charge 2.0 × 10–8 C each are joined by an insulating string of length 1m and the system is kept on a smooth horizontal table. Find the tension in the string.


Answer:

Given,


Charge on each particle,


Distance between the two charges,



The tension will adjust such that the net force is zero.



__________________________________________________


Formula used:


By Coulomb’s law, the electric force id given by:



Where ϵ0 is the permittivity of free space


q1 and q2 are the magnitude of charges


r is the distance of separation between the charges


(Here, q1 =q2 = q.)


__________________________________________________



Putting the values in the above formula, we get




Question 22.

Two identical balls, having a charge of 2.00 × 10–7 C and a mass of 100g, are suspended from a common point by two insulating strings each 50 cm long. The balls are held at a separation 5.0 cm apart and then released. Find

(a) the electric force on one of the charged balls

(b) the components of the resultant force on it along and perpendicular to the string

(c) the tension in the string

(d) the acceleration of one of the balls. Answers are to be obtained only for the instant just after the release.


Answer:

Given,


Mass,


Length of strings, l = 50cm = 0.5 m


Distance between spheres,


Magnitude of charge on each ball,


Let the magnitude of Tension be T. Let the magnitude of electric force between the spheres be Fe.


Note that tension will adjust itself so that there is no acceleration along its direction.



(a)


___________________________________________________


Formula used:


By Coulomb’s law, the electric force is given by:



Where ϵ0 is the permittivity of free space


q1 and q2 are the magnitude of charges


r is the distance of separation between the charges


(Here, )


___________________________________________________


The magnitude of the electric force, Fe is given by:




Hence, the electric force is 0.144 N along the line joining the charges and away from the other charge.


(b)


By trigonometry,




Now, along the string, there is component of acceleration. Hence,



Now, consider the direction perpendicular to the string,





Hence, the component of net force perpendicular to the string is 0.095N and away from the other charge (indicated by positive sign)


(c)From the free body diagram,



Rearranging to get the above expression, we get




(d)


Applying Newton’s second law along the y-direction,






Thus the acceleration of one of the balls is 0.95ms2 perpendicular to the string and going away from the other charge.



Question 23.

Two identical pith balls are charged by rubbing against each other. They are suspended from a horizontal rod through two strings of length20 cm each, the separation between the suspension points being 5 cm. In equilibrium, the separation between the balls is 3 cm. Find the mass of each ball and the tension in the strings. The charge on each ball has a magnitude 2.0 × 10–8 C.


Answer:

Given,


Length of strings,


Distance between suspension points,


Magnitude of charge on each ball,


Distance between the two balls,


Let the mass of each ball be m. Let the magnitude of Tension be T and the electric force between the balls be Fe.


To be in accordance with the fact, , the balls must be attracted to each other. Hence, they are oppositely charged.



By trigonometry,




At equilibrium, all forces must cancel out in accordance with newton’s first law. Hence,



___________________________________________________


Formula used:


By Coulomb’s law, the electric force is given by:



Where ϵ0 is the permittivity of free space


q1 and q2 are the magnitude of charges


r is the distance of separation between the charges


(Here, )


___________________________________________________


The magnitude of the electric force, Fe is given by:



Substituting values in (3), we get



From equation (1), we get




From equation (2), we get




Hence, mass of each of the balls is 8.2 grams and Tension in each of the ropes is 0.08N.



Question 24.

Two small spheres, each having a mass of 20g, are suspended from a common point by two insulating strings of length 40 cm each. The spheres are identically charged and the separation between the balls at equilibrium is found to be 4 cm. find the charge on each sphere.


Answer:

Given,


Mass,


Length of strings, l = 40cm = 0.4 m


Distance between spheres,


Let the magnitude of Tension be T. Let the electric force between the spheres be Fe. Let the magnitude of charge on each pitch ball be q.



By trigonometry,




At equilibrium, all forces must cancel out in accordance with newton’s first law. Hence,



___________________________________________________


Formula used:


By Coulomb’s law, the electric force is given by:



Where ϵ0 is the permittivity of free space


q1 and q2 are the magnitude of charges


r is the distance of separation between the charges


(Here, )


___________________________________________________


The magnitude of the electric force, Fe is given by:



From (2), we get,



From (3) and (1), we get







Question 25.

Two identical pith balls, each carrying a charge q, are suspended from a common point by two strings of equal length ℓ. Find the mass of each ball if the angle between the strings is 20 in equilibrium.


Answer:

Given:


Charges on ball = q


Angle between the balls = 2θ


Length of the strings = l


Let the magnitude of Tension be T. Let the electric force between the pitch balls be Fe. Let the mass of each pitch ball be m. The free body diagram is as follows:



Now, let the distance between the two charges be r.


By trigonometry,




By Coulomb’s law, the electric force id given by:



Where ϵ0 is the permittivity of free space


q1 and q2 are the magnitude of charges


r is the distance of separation between the charges


Here, q1 =q2 = q.


Now, let us find the magnitude of the electric force, Fe :



Substituting eq (1), we get



As the system is in equilibrium, all the forces in each direction must sum up to zero.



Substituting (2) in (3), we get



Substituting (4), we get



Substituting (5) in (6), we get




Question 26.

A particle having a charge of 2.0 × 10–4 C is placed directly below and at a separation of 10 cm from the bob of a simple pendulum at rest. The mass of the bob is 100g. What charge should the bob be given so that the string becomes loose?


Answer:


Given:
Charge of the particle : q = 2.0 × 10–4 C
Distance between charged particle and the bob:
r =10 cm=0.1m
Mass of the bob : m = 100 g = 100 × 10-3 kg.
Formula used:
T = mg
Where T is the tension in the string and g is the acceleration due to gravity.
∴ T = 0.1 × 9.8
∴ T = 0.98 N
Now let the charge on the bob be : q’
Now the electrostatic force between the bob and the particle is given as:
This is the Coulomb’s Law. Where ϵ0 is the permittivity of free space and it’s value is : ϵ0 = 8.85418782 × 10-12 m-3 kg-1 s4 A2,q is the charge of the particle and q’ is the charge of the bob, r is the distance between charged particle and the bob.
Now, in order to have a loose string the tension in the string should be zero: T=0
For tension to be zero, the particle must repel the bob along the direction of the tension.
Also, weight of the bob is in opposite direction to the string.
Thus the equation is:
T + F = mg


Hence, the charge on the bob should be 5.4 × 10-4 C so that the string would become loose.



Question 27.

Two Particles A and B having charges q and 2q respectively are placed on a smooth table with a separation d. A third particle C is to be clamped on the table in such a way that the particles A and B remain at rest on the table under electrical forces. What should be the charge on C and where should it be clamped?


Answer:


Given:
Charge of the particle A : q
Charge of the particle B : 2q

Distance between A and B : d
Formula used:
We will be using Coulomb’s Law:
Where ϵ0 is the permittivity of free space and it’s value is : ϵ0 = 8.85418782 × 10-12 m-3 kg-1 s4 A2 and F is the electrostatic force between two charges. And r is the distance between two charges.
The diagram below shows the given conditions:

C is the third particle clamped on the table. Let charge of point C be q’
For A and B to remain at rest, the electrostatic forces from A and B must cancel out each other. Which means magnitude of force from A to C : should be equal and opposite to the force from B to C : as Point B has twice the charge of A.
Thus the equation of rest is:


Here x is the distance between charge A and C.
From Coulomb’s Law,



Substituting we get,
Since we know that the magnitudes of FAC and FBC the forces are equal,








Rationalizing the denominator we get,
Substituting value of x in the equation (1) we get,







Hence, the charge on the C is -q(6-4√2) C and it should be clamped at a distance of (√2-1)d .



Question 28.

Two identically charged particles are fastened to the two ends of a spring of spring constant 100 N m–1 and natural length 10 cm. The system resets on a smooth horizontal table. If the charge on each particle is 2.0 × 10–8 C, find the extension in the length of the spring. Assume that the extension is small as compared to the natural length. Justify this assumption after you solve the problem.


Answer:


Given:
Natural Length of the spring: l = 10cm = 0.1m
Spring Constant : K = 100 N m–1
Charge on each particle: q = 2.0 × 10–8 C
Separation between two charges = l
Formula used:
Let the extension be ‘x’ m.
We use Coulomb’s Law:
Where Fe is the electrostatic force, k is a constant .
k = = 9× 109 Nm2C-2 and r is the distance between two charges.
Since the electrostatic force is repulsive in nature, the spring will exert a restoring spring force Fr.
F= -Kx

Here, K is the spring constant and x is the extension.
Negative sign is because the restoring spring force us opposite to the applied force. The system would be in equilibrium when the Electrostatic force of repulsion between the two charges is equal to the spring force

Fe + Fr = 0





Yes, the assumption is justified. When two similar charges are present at two ends, they will exert repulsive force on each other. The spring will extend due to its elastic nature.
The repulsive force will have an opposite force called as restoring force in the spring. This force is directly proportional to the extension of the spring and depends on the elasticity if the material. If the extension is large compared to the natural length, then the restoring force would be proportional to the high powers of the extension.



Question 29.

A particle A having a charge of 2.0 × 10–6 C is held fixed on a horizontal table. A second charged particle of mass 80g stays in equilibrium on the table at a distance of 10 cm from the first charge. The coefficient of friction between the table and this second particle is μ = 0.2. Find the range within which the charge of this second particle may lie.


Answer:

Given:
Charge of the particle A : q1 = 2.0 × 10–6 C.
Mass of the second charged particle i.e. B:
m=80 g=80×10-3 kg.
Separation between both the charged particles:
r = 10 cm = 0.1 m
Coefficient of friction between the table and this second particle: μ = 0.2

Formula used:
Let the charge of the particle B be q2
We use Coulomb’s law:
Where Fe is the electrostatic force on b due to A, k is a constant .
k = = 9× 109 Nm2C-2 and r is the distance between two charges.
The Friction force is given as:
Fr = μN= μmg
Here, μ is the coefficient of friction, N is the normal reaction of table to the particle. N=mg.
It is given that particle B is at equilibrium with the table,
Thus
Fe = Fr



Hence, the range within which the charge of the second particle lies is ± 8.71× 10-8 C. Since, charge can be positive or negative, thus ± .



Question 30.

A particle A having a charge of 2.0 × 10–6 C and a mass of 100g is placed at the bottom of a smooth inclined plane of inclination 30°. Where should another particle B, having same charge and mass, be placed on the incline so that it may remain in equilibrium?


Answer:


Given:
Charge of the particle A and B: q1=q2= q = 2.0 × 10–6 C.
Mass of the particles A and B : m = 100 g =0.1 kg.
Inclination of the inclined plain : θ = 30°

Formula used:
The particle A and B will exert electrostatic repulsive forces on each other.
This can be given by Coulomb’s Law:
We use Coulomb’s law:
Where Fe is the electrostatic force on b due to A, k is a constant .
k = = 9× 109 Nm2C-2 and r is the distance between two charges.
Particle B would face a friction force opposite to the rolling force due to inclination. It is given as:
F= mgsinθ
For particle B to remain in equilibrium with the inclination and particle A. The electrostatic force of repulsion and the friction force must be equal.
Fe = F




Hence, particle b must be place at a distance of 0.2701m from particle A to remain in equilibrium.



Question 31.

Two particles A and B, each having a charge Q, are placed a distance d apart. Where should a particle of charge q be placed on the perpendicular bisector of AB so that it experiences maximum force? What is the magnitude of this maximum force?


Answer:


Given:
Charge on particles A and B: q1= q2= Q
Separation between A and B : d

here ‘x’ is the distance at which particle C of charge ‘q’ is place on the perpendicular bisector of AB.
Formula used:
From the figure we can find sinθ :
The horizontal components of force cancel each other.
Total vertical component of force is: F’= 2Fsinθ
We use Coulomb’s Law:
Where F is the electrostatic force on b due to A, k is a constant .
k = = 9× 109 Nm2C-2 and r is the distance between the two charges.
Here r =
Substituting we get,












The magnitude of maximum Force is:

Hence, the particle C must be placed at a distance of from the perpendicular bisector of AB so as to experience a maximum force of magnitude



Question 32.

Two particles A and B, each carrying a charge Q, are held fixed with a separation d between them. A particle C having mass m and charge q is kept at the middle point of the line AB.

(a) If it is displaced through a distance x perpendicular to AB, what would be the electric force experienced by it.

(b) Assuming x << d, show that this force is proportional to x.

(c) Under what conditions will the particle C execute simple harmonic motion if it is released after such a small displacement?

Find the time period of the oscillations if these conditions are satisfied.



Answer:


Given:
Charge of particles A and B : q = Q
Separation between A and B : d
(a)

here ‘x’ is the distance at which particle C of mass m and charge ‘q’ is place on the perpendicular bisector of AB.
Formula used:
From the figure we can find sinθ :
The horizontal components of force cancel each other.
Total vertical component of force is: F’= 2Fsinθ
We use Coulomb’s Law:
Where F is the electrostatic force on b due to A, k is a constant .
k = = 9× 109 Nm2C-2 and r is the distance between the two charges.
Here r =
Substituting we get,

F’ is the net electric force experience by particle C of charge q.
(b)
When x ≪ d,
Thus, x2 can be neglected.
Substituting we get,


Thus, force is proportional to x.
(c)
The condition for Simple harmonic Motion of a particle is:
Here, m is the mass of the particle C, ω is the angular frequency.
Thus comparing two equations of F’, we get
We know that Where t is the Time Period.




Hence time period when the particle is released after a small displacement under SHM is



Question 33.

Repeat the previous problem if the particle is displaced through a distance x along the line AB.


Answer:


Given:
Charge of particles A and B : q = Q
Separation between A and B : d

Here x is the displacement of particle C along AB.
Distance between A and C is :
Distance between B and C is :
Formula used:
For part (a), we will be using Coulomb’s Law,
Where F is the electrostatic force on B due to A, k is a constant .
k = = 9× 109 Nm2C-2 and r is the distance between the two charges.
q1=q2=Q.
now, the net force acting on C is:





Fnet is the net electric force experience by particle C of charge q.
(b)
When x≪d,
x2≪(d/2)2
Thus x2 can be neglected.
We get,

∴ Fnet α x

Thus, when x≪d force is proportional to x

(c)
The condition for Simple harmonic Motion of a particle is:
F'=mω2x
Here, m is the mass of the particle C, ω is the angular frequency.
Let Fnet = F’
Thus comparing two equations of F’, we get
We know that Where t is the Time Period.


Hence time period when particle is displaced along AB is




Question 34.

The electric force experienced by a charge of 1.0 × 10–6 C is 1.5 × 10–3 N. Find the magnitude of the electric field at the position of the charge.


Answer:


Given:
Electric Force exerted by a charge: F = 1.5 × 10–3 N
Charge: q = 1.0 × 10–6 C
Formula used:
Here we use:
Where, F is the electric force, q is the charge and E is the electric field at the position of the charge.
Substituting we get,



Hence, the magnitude electric field at the position of the charge is 1.5× 103 N/C



Question 35.

Two particles A and B having charges of +2.00 × 10–6 C and of –4.00 × 10–6 C respectively are held fixed at a separation of 20.0 cm. Locate the point(s) on the line AB where (a) electric field is zero (b) the electric potential is zero.


Answer:


Given:
Charge on particle A: qA = +2× 10-6 C
Charge on particle B : qB = -4× 10-6 C
Separation between A and B : r = 20.0 cm=0.2m

Formula used:
Electric Field given as:
Here, k is a constant and k= =9× 109 Nm2C-2 . q is the point charge and r is the distance between two charges.
Also,
Electric potential is given as:
Here, k is a constant and k= =9× 109 Nm2C-2 .q is the point charge and r is the distance between two charges.
Now,

Let there be a point at distance ‘x’ from A where net electric field is zero.
Distance between A and the point be : rA = x m
Distance between B and the point be : rB = (0.2-x) m










Hence, electric field is zero at a point 0.4825 m from A along AB.
Now,
Zero Net potential at the point is








OR

Hence, potential can be zero at 0.0666 m from A along AB and at 0.2 m from B along AB.



Question 36.

A point charge produces an electric field of magnitude 5 N C–1 at a distance of 40 cm from it. What is the magnitude of the charge?


Answer:


Given:
Electric field due to a point charge: E = 5 N C–1
Distance between the point charge and the point at which electric field is produced : r = 40 cm = 0.4 m
Formula used:

Electric field is given as:
E=kq r2
Here , k is a constant and k= = 9× 109 Nm2C-2 . q is the point charge .
Substituting we get,


Hence, magnitude of the charge is 8.89× 10-11 C.



Question 37.

A water particle of mass 10.0 mg and having a charge of 1.50 × 10–6 C stays suspended in a room. What is the magnitude of electric field in the room? What is its direction?


Answer:


Given:
Mass of the water particle : m= 10.0 mg = 10× 10-6 kg
Charge of the particle: q = 1.5× 10-6 C

The suspended particle will be under the influence of gravity.
Hence , gravity will act in downward direction as shown in the figure.
Formula used:
We know that,
Fe = qE
Where F is the electrostatic force, q is the charge and E is the electric field produced by the charge.
But, Force due to gravity : is FG= mg
Here, m is the mass of the particle and g is the acceleration due to gravity.
Now, for the particle to stay suspended in the room, the downward gravitational force must be equal and opposite to the electric force.
∴Fe=FG

∴ qE=mg


Thus, the magnitude of electric field due to the charged water molecule suspended in the room is 65.33 NC-1 and it is in upwards direction opposite to the gravitational force.



Question 38.

Three identical charges, each having a value 1.0 × 10–8 C, are placed at the corners of an equilateral triangle of side 20 cm. Find the electric field and potential at the center of the triangle.


Answer:


Given:
Value of three identical charges: q = 1.0× 10–8 C
Side of the equilateral triangle: l = 20cm = 0.2m

From the diagram,


A,B and C are the three vertices having equal charge q.
EA,EB and Ec are the electric fields at the center of the triangle due to charges A,B and C respectively.
h is the height of the equilateral triangle
and r is the distance from the center of the triangle to it’s all three vertices.



Formula used:
Formula for potential at a point is:


Where k is a constant and k= =9× 109 Nm2C-2 .q is the point charge and r is the distance between the centre of the triangle and the vertex.


Since charges are equal at A,B and C:
The Field from B and C are resolved into horizontal and vertical components as seen from the figure.
Here θ is 30° as every angle of an equilateral triangle is 60o .
The horizontal components balance each other.
Therefore net electric field ,
Enet = EA -(EB sinθ+ECsinθ)
∴Enet= E-(Esinθ+Esinθ)
∴Enet = E(1-sin(30)-sin(30) )
∴Enet =E(1-0.5-0.5)
∴Enet =0
Thus, the electric field at the center of the given equilateral triangle is zero.
Now, using Pythagoras theorem to find value of h,


We know that, in an equilateral triangle
Thus we get,


Since Electric field is same for all three points: VA=VB=VC
The potential at the center is :




Hence , potential at the centre of the triangle is 2341 V and Electric field at the center is zero.



Question 39.

Positive charge Q is distributed uniformly over a circular ring of radius R. A particle having a mass m and a negative charge q, is placed on its axis at a distance x from the centre. Find the force on the particle. Assuming x <<R, find the time period of oscillation of the particle if it is released from there.


Answer:


Given:

Charge on the ring: Q
Radius of the ring : R
Charge of the particle at point P : q
Mass of the particle : m
Distance of P from the centre of the ring: x
Distance of P from the element A : l
Formula used:
Electric force is given as:
Where F is the electric force, q is the charge and E is the electric field.

Newton’s Law gives :
Time period is given as:
Where, l is the length. In this case OP=x and a is the acceleration.



Where F is the electric force, q is the charge and E is the electric field.
Consider an element of charge dQ on the ring at A.
Electric field at P due to the element A is given as:
Here, dE is the electric field due to element A. k= =9× 109 Nm2C-2, dQ is the charge of the element A and l is the distance between A and P.
Now,
And

Ecosθ

Therefore, the net electric field at P due to the entire ring is:



Therefore substituting in equation (1) we get,
Here, F is the electric force on the particle due to entire charged ring.

Now the condition given in the question is x≪R. Thus, x2 can be neglected.

Hence using equation (2) and (3) we get,
Putting the value of k:

Hence time period of oscillation of the particle is




Question 40.

A rod of length L has a total charge Q distribute uniformly along its length. It is bent in the shape of a semicircle. Find the magnitude of the electric field at the centre of curvature of the semicircle.


Answer:


Given:
Length of the rod: L
Uniformly distributed charge: Q
When the rod is bent into a semicircle, let R be its radius.

Here, dθ is the width of the angular element on the circumference. The displacement of the element with width dθ is Rdθ. Let the displacement of the element be dl.
The Electric field is divided in to horizontal and vertical components. Horizontal components are cancelled out.
Formula Used:
We know that:
Here, λ is the linear charge density of the rod, Q is the charge of the rod and L is the length of the rod.
for a charge dq of element dl we have
We know what dl =rdθ
The formula for electric field is:
Here, k is a constant and k= =9× 109 Nm2C-2 . q is the point charge and r is the distance between the charge and the point of influence.
here r=R.
The net electric field due to vertical component is:
Here, dE is the electric field due to element dl having charge dq and the limits of the integral would be from 0 to π as it is a semicircle.






If we substitute value of k we get,


As L=π R.
Hence electric field at the centre of the curvature is



Question 41.

A 10 cm long rod carries a charge of +50 μC distributed uniformly along its length. Find the magnitude of the electric field at a point 10 cm from both the ends of the rod.


Answer:


Given:
Length of the rod : L = 10cm = 0.1m
Charge on the rod; q = +50 μC= 50× 10-6 C



Here, C is 10 cm = 0.1 m away from both ends of the rod AB.
Distance between C and centre of AB : r
Formula used:
From Pythagoras Theorem,
r2+(L/2)2=(BC)2


∴ r2= [0.1]2- [0.05]2


∴ r2= 7.5×10-3


∴r=√ (7.5×10-3)


∴r=0.0866 m
Now we know that, Electric field at a point on the perpendicular bisector of a uniformly charged rod is:
Here, k is a constant and k= =9× 109 Nm2C-2 . q is the point charge, L is the length of the rod and Q is the magnitude of the charge.
Substituting we get,

Hence, electric field at a point 10 cm away from the ends of the rod is 5.2 × 107 NC-1.



Question 42.

Consider a uniformly charged ring of radius R. Find the point on the axis where the electric filed is maximum.


Answer:


Given:
Charge of the ring: Q
Radius of the ring :R
Let P be the point where electric field is found.
Distance between center of the ring and P is x.

Formula used:
We know that electric field at any point on the axis at a distance x from the center is:
Where k is a constant and k= =9× 109 Nm2C-2. Q is the charge of the ring, x is the distance between center of the ring and point P and R is the radius of the ring.
Now for the electric field to be maximum, we use the maxima property:
Taking derivative of E w.r.t x,






Hence Electric field is maximum at on the axis.



Question 43.

A wire is bent in the form of a regular hexagon and a total charge q is distributed uniformly on it. What is the electric field at the center? You many answer this part without making any numerical calculations.


Answer:


Given:
Uniformly distributed Charge on the regular hexagon : q

When a wire is bent in the form of a regular hexagon having charge q uniformly distributed, each point on the hexagon will contribute same magnitude of electric field.
Say, 6 vertices of the hexagon have same charge q.
Thus they will produce same electric field E at the center.
This electric field gets nullified as same magnitude is acting from both sides.
Formula used:
Electric field at a point due to a point charge is given as:
Where k is a constant and k= =9× 109 Nm2C-2. q is the charge and r is the distance between the point and the charge.
Mathematically:
E1=E2
From the figure E1 is the electric field due to vertex 1 at the center and E2 is the electric field due to vertex 2 at the center diametrically opposite to vertex 1.
Thus net electric field at center due to 1 and 2 is
Enet = E1-E2

∴ Enet = 0

Eventually electric field due to all 6 vertices cancel out each other at the center.
Hence, electric field at the center of the regular hexagon is zero.



Question 44.

A circular wire-loop of radius a carries a total charge Q distributed uniformly over its length. A small length dL of the wire is cut off. Find the electric field at the center due to the remaining wire.


Answer:


Given:
Radius of the circular loop : a
Total charge on the wire : Q
Length of the cut off wire: dL
Formula used:
Electric field is given as:
Here, k is a constant and k= =9× 109 Nm2C-2 . q is the point charge and r is the distance between the charge and the point of influence.
Here r=a=radius of the loop.

We know that, electric field at the center of the uniformly charged circular wire

Which means that sum of electric field due to cut off wire and remaining wire is zero.
We know that,
Where λ is the linear charge density. Q is the total charge and L is the length of the wire.
Charge on the element dL be dq:


Here L = 2πa=circumference. Here, a is the radius of the loop and L is the Length of the loop.

Thus, electric field due to dL(cutoff wire) at the center is:


Since


Thus, magnitude of electric field at the center of the circular wire due to remaining wire is E= but in opposite direction to that of the field due to cut off wire.



Question 45.

A positive charge q is placed in front of a conducting solid cube at a distance d from its center. Find the electric field at the center of the cube due to the charges appearing on its surface.


Answer:


Given:
Charge placed in front of a solid cube: +q
Distance between the charge and the center of the cube: d

Formula used:
Formula for electric field is:
Here, k is a constant and k= =9× 109 Nm2C-2 . q is the point charge and r is the distance between the charge and the point of influence.
Since we have to find electric field at the center of the cube


Hence, electric field at the center of the cube due to a positive charge at a distance d from the center of cube is
.



Question 46.

A pendulum bob of mass 80 mg and carrying a charge of 2 × 10–8 C is at rest in a uniform, horizontal electric field of 20 kVm–1. Find the tension in the thread.


Answer:


Given:
Mass of the bob: m = 80 mg= 80× 10-6 kg
Charge on the bob: q = 2 × 10–8 C
Electric field : E = 20 kVm–1 = 20× 103 Vm-1.

Here the electric field is horizontal.
The tension in the string is resolved into two components:
Horizontal component : Tsinθ
Vertical component : Tcosθ
As shown in the figure.
Formula Used:
Since electric field acts on the bob having charge q, it experiences an electric field in horizontal direction.
Electric force is given as:
F=qE
F is electric force, q is the charge on the bob and E is the horizontal electric field. This electric force is balanced by the horizontal component of the tension.
Tsinθ = qE
Now the vertical component of tension is balanced by the weight of the bob.
Tcosθ=mg
Dividing we get,




Substituting the value of θ , we can calculate tension in the string.


Hence the tension in the string is 8.8×10-4 N.



Question 47.

A particle of mass m and charge q is thrown at a speed u against a uniform electric field E. How much distance will it travel before coming to momentary rest?


Answer:


Given:
Mass of the particle: m
Charge of the particle: q
initial velocity of the particle when thrown : u
Uniform electric field : E
Formula used:
When a charge q moves in an uniform electric field E, it experiences an electric force F.
F=qE
But this particle is thrown against the electric field, hence the force experience would be negative.
F=-qE
According to Newton’s Second Law:
F=ma
Where a is the acceleration and m is the mass of the body.
Thus,
We will be using one of the equations of motion.
Here, v is the final velocity of the particle, u is the initial velocity, a is the acceleration and s is the displacement.
Since particle comes to rest, final velocity: v=0
Thus,


Hence, the particle will travel units before coming to rest.



Question 48.

A particle of mass 1g and charge 2.5 × 10–4 C is released from rest in an electric field of 1.2 × 104 N C–1.

(a) Find the electric force and the force of gravity acing on this particle. Can one of these forces be neglected in comparison with the other for approximate analysis?

(b) How long will it take for the particle to travel a distance of 40 cm?

(c) What will be the speed of the particle after travelling this distance?

(d) How much is the work done by the electric force on the particle during this period?



Answer:


Given:
Mass of the particle : m= 1g = 10-3 kg
Charge of the particle: q = 2.5 × 10–4 C
Electric Field: E = 1.2 × 104 N C–1
Distance travelled : s = 40 cm = 0.4 m
Initial velocity: u = 0
Formula used:

(a)
When a charge q moves in an uniform electric field E, it experiences an electric force Fe.
Fe = qE
substituting the values we get,
∴ Fe = 3N
Force of gravity would be:
Where m is the mass of the particle and g is the acceleration due to gravity.

Since the mass of the particle is very low, force due to gravity can be neglected.
(b)
From Newton’s Second Law,
F=ma
where a is the acceleration of the body and m is the mass.
acceleration of the particle is:


Here, s is the distance travelled, u is the initial velocity, t is the time required to travel s, and a is the acceleration of the particle.
Substituting we get,



It will take 0.0163 seconds for the particle to travel a distance of 40 cm.
(c)
Using another equation of motion
Here v is the final velocity of the particle,u is the initial velocity of the particle, a is the acceleration and s is the distance travelled by the particle.
Substituting we get,


After travelling 40cm the speed of the particle will be 48.9 m/s.
(d)
We know that,
Work=Force×Displacement
Thus work done by the electric force:
W = F × s
∴ W = 3× 0.4
∴ W = 1.2 J

Hence , work of 1.2 J is being done by the electric force on the particle.



Question 49.

A ball of mass 100g and having a charge of 4.9 × 10–5 C is released from rest in a region where a horizontal electric field of 2.0 × 104 C NC–1 exists.

(a) Find the resultant force acing on the ball.

(b) What will be the path of the ball?

(c) Where will the ball be at the end of 2s?



Answer:


Given:
Mass of the ball : m = 100 g = 100× 10-3 kg =0.1 kg
Charge on the ball : q = 4.9 × 10–5 C
Horizontal electric field : E= 2.0 × 104 C NC–1
Initial Velocity of the ball: u = 0

Here, R is the resultant Force due to gravitational force Fg and electric force Fe.
Formula used:
(a)
Electric force Fe due to charge q and electric field E is
Fe = qE
Gravitational force Fg experience due to mass of the ball m and acceleration due to gravity g is:
Fg=mg

Thus we see that Fe = Fg
The Resultant force R can be calculated by:
R2=Fe2+Fg2

∴ R2= (0.98)2+(0.98)2


∴R=√1.9208


∴R=1.3859 N
Hence the resultant force of 1.3859 N is acting on the ball.
(b)
We take tangent of the angle θ

But Fg= Fe,

Hence, the path of the ball is along a straight line and inclined at an angle of 45° with the horizontal electric field.
(c)
Here, we will be using one of the equations of motion.
Here s is the distance covered by the ball, u is the initial velocity of the ball, a is the acceleration of the ball and t is the time required to cover s.
We need to find s at t=2s
Firstly ,vertical displacement due to gravitational force is:a=g

Secondly,
Horizontal displacement due to electric force is:


Thus net displacement = (sv2+ sh2)1/2
∴ Net displacement = ((19.6)2+(19.62))1/2
∴ Net displacement = 27.71 m


Thus, the ball will be at a distance of 27.71m after 2s



Question 50.

The bob of a simple pendulum as a mass of 40g and a positive charge of 4.0 × 10–6 C. It makes 20 oscillations in 45s. A vertical electric field pointing upward and the magnitude 2.5 × 104 N C–1 is switched on. How much time will it now take to complete 20 oscillations?


Answer:


Mass of the bob: m = 40 g = 0.04 kg
Charge of the bob: q = 4.0 × 10–6 C
Oscillations in t=45s : 20
Vertical electric field: E = 2.5 × 104 N C–1.
Formula used:
Time period of a simple pendulum is:
here l is the length of the string .
Thus, when there is no electric field, the time period is:
When a vertical electric field is applied, the positively charged bob will experience a vertical acceleration due to electric force. Thus net acceleration of the bob = g-a
Time period of the simple pendulum when vertical electric field applied is:
Now,
Since a is due to electric force, by Newton’s Second Law and Formula for electric force, we get



Thus g-a = 9.8-2.5 = 7.3 m/s2
Taking ratio we get:






Hence time required by the bob to complete 20 oscillations in presence of a vertical electric field is 52 seconds.



Question 51.

A block a mass m having a charge q is placed on a smooth horizontal table and is connected to a wall through an unstressed spring of spring constant κ as shown in figure. A horizontal electric field E parallel to the spring is switched on. Find the amplitude of the resulting SHM of the block.



Answer:


Given:
Mass of the block: m
Charge of the block: q
Formula used:
When a charged body of charge q and mass m is brought under an horizontal electric field E, it will experience electric force Fe in the direction of the field.
Fe=qE
When the block is accelerated due to the electric force, the spring will cause a restoring spring force in the opposite direction.
Spring force is:
Fs = -kx
Where, k is the spring constant and x is the distance the spring is stretched or compressed.
Here x is the amplitude.
Thus,
Fe = Fs
∴ qE = -kx
We used modulus as the amplitude cannot be negative.
Hence, is the amplitude of the resulting SHM of the block.



Question 52.

A block of mass m containing a net positive charge q is placed on a smooth horizontal table which terminates in a vertical wall as shown in figure. The distance of the block from the wall is d. A horizontal electric field E towards right is switched on. Assuming elastic collisions (if any) find the time period of the resulting oscillatory motion. Is it a simple harmonic motion?



Answer:


Given:
Mass of the block : m
Charge on the block: q
Distance between block and the wall : d
Horizontal electric field: E
Initial velocity: u=0


Formula used:

It is not Simple Harmonic Motion. In SHM the acceleration is directly proportional to the displacement of the body and in opposite direction of the displacement. In this case, acceleration is proportional to the displacement but not in the opposite direction.
We know that,


Here, F is the horizontal Electric force , q is the charge of the body and E is the electric force (horizontal), a is the acceleration and m is the mass of the body
To find time required by the block to collide into the wall can be calculated by laws of motion.
Here s is the displacement; s=d. u is the initial velocity , a is the acceleration of the block and t is the time required to travel the displacement.

This is the time required by the block to travel d and hit the wall.
Since it is assumed to be elastic collision , after collision with the wall the block will take same time ‘t’ to come back till it’s velocity is zero.
Thus Total time take:T

Time period of the resulting oscillatory motion is



Question 53.

A uniform electric field of 10 N C–1 exists in the vertically downward direction. Find the increase in the electric potential as one goes up through a height of 50 cm.


Answer:


Given:
Electric field: E= 10 N C–1
Change in Height : dh = 50 cm= 0.5 m
Formula used:
Since the electric field exists in vertically downward direction, E becomes negative as one goes up. E= -10 N C–1
Change in potential is :
dV = -E.dr
Here, dV is change in potential, dr is the distance moved and E is the electric field.
In this case dr=dh
∴ dV = -(-10)× 0.5
∴ dV = 5 V
Hence, Increase in potential as one goes up by 50cm is 5V.



Question 54.

12 J of work has to be done against an existing electric field to take a charge of 0.01 C from A to B. How much is the potential difference VB – VA?


Answer:


Given:
Work done against electric field: W= 12 J
Charge : q= 0.01 C
Formula used:
Work done is given by:
W=Potential difference×charge

∴W=(VB-VA )×q
Here, VB-VA is the potential difference and q is the charge.

Hence, potential difference is 1200 V



Question 55.

Two equal charges, 2.0 × 10–7 C each, are held fixed at a separation of 20 cm. A third charge of equal magnitude is placed midway between the two charges. It is now moved to a point 20 cm from both the charges. How much work is done by the electric field during the process?


Answer:


Given:
Magnitude of all three charges: q1=q2=q3= q= 2.0 × 10–7 C
Separation between first two charges: d = 20 cm = 0.2m

From the figure,
Distance between charge at A and both the charges:
r =10cm= 0.1 m
Distance between charge at b and both the charges:
r’= 20 cm= 0.2 m
Formula used:
Potential is given as
Here, Here, k is a constant and k= =9× 109 Nm2C-2 . q is the point charge and r is the distance between the charge and the point of influence.
Since A is midway between two point charges, potential at A will be due to both the charges:


Now, when charge at A is displaced to B, potential difference is created. Potential at B is due to both the charges at 0.2 m equally from B.
Potential at B:


Thus, Potential difference is:
VA-VB =36000-18000
∴ VA-VB = 18000
Work done is:
W= (VA-VB) × q
∴ W = 18000 × 2.0 × 10-7
∴ W = 3.6 × 10-3 J

Hence, 3.6× 10-3 J of work is being done by the electric field during whole process.



Question 56.

An electric field of 20 N C–1 exists along the x-axis in space. Calculate the potential difference VB – VA where the points A and B are given by,

(a) A = (0, 0); B = (4m, 2m)

(b) A = (4m, 2m); B = (6m, 5m)

(c) A = (0, 0); B = (6m, 5m)
Do you find any relation between the answers of parts (a), (b) and (c)?



Answer:


Given:
Magnitude of Electric field: E = 20 N C–1
E is along x-axis
Formula used:
As Electric field is along x-axis, potential difference will be along x-direction. Which means only x co-ordinates will be considered.
We know that,
Here dV is the change in potential : dV= VB-VA
E is the electric field along positive x axis and ds is the change in displacement.
(a)
A = (0, 0); B = (4m, 2m)
∴ VB-VA= -20×(4-0) = -80 V
(b)
A = (4m, 2m); B = (6m, 5m)
∴ VB-VA= -20×(6-4) = -40 V
(c)
A = (0, 0); B = (6m, 5m)
∴ VB-VA= -20×(6-0) = -120 V
(d)
From (a),(b) and (c), we conclude that:
Potential difference of at points A = (0, 0), B = (6m, 5m)
= Potential difference at points A = (0, 0), B = (4m, 2m)
+ Potential difference at points A = (4m, 2m), B = (6m, 5m)



Question 57.

Consider the situation of the previous problem. A charge of –2.0 × 10–4 C is moved from the point A to the point B. Find the change in electrical potential energy UB – UA for the cases (a), (b) and (c).


Answer:


Given:
Magnitude of Electric field: E = 20 N C–1
Magnitude of charge moved from A to B = –2.0 × 10–4 C
Formula used:
Change in Electrical potential energy is
Here, ΔU is change in Electrical potential energy, Δ V is change in potential and q is the charge displaced.
ΔU = UB-UA where UB is electric potential energy at B and UA is electric potential energy at A
(a)
A = (0, 0); B = (4m, 2m)
∴ VB-VA= -20×(4-0) = -80 V
Thus Change in Electrical potential energy is
UB-UA = (VB – VA)× q
∴ Δ U = -80 × -2.0× 10-4
∴ Δ U = 0.016 J
(b)
A = (4m, 2m); B = (6m, 5m)
∴ VB-VA= -20×(6-4) = -40 V
Thus Change in Electrical potential energy is
UB-UA = (VB – VA)× q
∴ Δ U = -40 × -2.0× 10-4
∴ Δ U = 0.008 J

(c)
A = (0, 0); B = (6m, 5m)
∴ VB-VA= -20×(6-0) = -120 V
Thus Change in Electrical potential energy is
UB-UA = (VB – VA)× q
∴ Δ U = -120 × -2.0× 10-4
∴ Δ U = 0.024
Hence, for the cases (a),(b) and (c) the change in electric potential energy when the charge is moved from A to B is 0.016 J,0.008 J , 0.024 J respectively.



Question 58.

An electric field exists in the space. If the potential at the origin is taken to be zero, find the potential at (2m, 2m).


Answer:


Given:
Electric field: = 20+ 30
V(0,0) = 0
Final position:
Formula used:
We know that Electric potential is given as:
Where ds is the change in displacement.
In vector form
Here, r̅ is the changed position from origin.



Hence potential at (2m,2m) is -100 V.



Question 59.

An electric field exists in the space, where A = 10 Vm–2. Take the potential at (10m, 20m) to be zero. Find the potential at the origin.


Answer:


Given:
Electric Field :
A = 10 Vm-2
V(10m,20m) = 0
Formula used:
Change in potential is:
Where dV is change in potential, E is the electric field and dx is the change in displacement.
In vector form:

Integrating we get:


Here limits 10 to 0 is taken as 10 m is the x co-ordinate and origin has (0,0)
Hence. potential of 500 V exists at the origin.



Question 60.

The electric potential existing in space is
V(x, y, z) = A (xy + yz + zx).

(a) Write the dimensional formula of A.

(b) Find the expression for the electric field.

(c) If A is 10 SI units, find the magnitude of the electric filed at (1m, 1m, 1m).



Answer:


Given:
Electric Potential :V(x,y,z) = A (xy + yz + zx).
A = 10 SI units.
Formula used:
(a)
From the given data V(x,y,z) = A (xy + yz + zx), we can say that Voltage is the product of A and length2(xy+yz+zx)
Volt = A× m2We know that Volt = , where Q=

Thus, dimensions of A are
(b)
Formula for electric field is:
Where


Hence , expression of electric field is
(c)
A = 10 SI units
(x,y,z) = (1m,1m,1m)
Substituting in the expression of E we get:

Magnitude of Electric field is



Hence, magnitude of Electric field at (1m,1m,1m0 is 34.64 NC-1.



Question 61.

Two charged particles, having equal charges of 2.0 × 10–5 C each, are brought from infinity to within a separation of 10 cm. Find the increase in the electric potential energy during the process.


Answer:


Given:
Charge of two particles: q1=q2= 2.0 × 10–5 C
Separation between two charges: r = 10 cm = 0.1 m
Formula used:
Electric potential energy is given as:
Where,k is a constant and k= =9× 109 Nm2C-2 . q is the point charge and r is the separation between two charges.
When two charges were at infinity, the separation between them was infinite
Thus,

Now, when the separation between them was 10 cm;
Where Uf is the final electric potential energy.
Substituting the values we get,

Now, increase in electric potential energy: Δ U


Hence, Electric potential energy increased by 36 J during the process.



Question 62.

Some equipotential surfaces are shown in figure. What can you say about the magnitude and the direction of the electric field?



Answer:


Given:
From figure (a)
Angle between equipotential surfaces and the displacement x
:θ = 30°
Change in potential : dV = 10 V
Change in displacement between two consecutive equipotential surfaces: dx = 10 cm = 0.1 m
From figure (b)
Increase in radius from center: dr= 10cm=0.1m
Formula used:
(a)
As we know that the electric field is always perpendicular to the equipotential surface

From the above diagram, the angle between Electric field and dx ; θ’ = 90° +30° =120°
Change in electric potential is given as:




Hence the magnitude of electric field is 200 V/m making an angle of 120° with the x axis.
(b)
As we know that the electric field is always perpendicular to the equipotential surface

Radius increases by: dr= 10 cm = 0.1 m
As is perpendicular to the equipotential surface, and dr would be along same line as shown in the figure above.
Thus angle between and dr : θ = 0

We know that potential at a point due to a charge q is given as:
Where, k is a constant and k= =9× 109 Nm2C-2 . q is the point charge and r is radius of that surface.
Consider potential at point A where r=0.1 m


Electric field is given as:
Substituting value of kq we get,
Hence, the magnitude of the electric field is and it’s direction is radially outward with decreasing with increasing radii.



Question 63.

Consider a circular ring of radius r, uniformly charged with linear charge density λ. Find the electric potential at a point on the axis at a distance x from the center of the ring. Using this expression for the potential, find the electric field at this point.


Answer:


Given:
Radius of the circular ring: r
Linear charge density : λ
Distance of a point from the center of the ring : x

From the diagram we can see that,
Point P is at a distance x from the center of the ring.
Point P is at a distance of from the surface of the ring: r’ =
Circumference of the ring is : L = 2π r
Formula used:
We can see that, Electric field as p is resolved into vertical and horizontal components. As the ring is symmetric, vertical components are cancelled out and horizontal components add.
Thus Enet =Ecosθ , where θ is the angle between x and .
We know that,
Where, λ is the linear charge density, Q is the Total charge due to whole ring and L is the circumference of the ring.
Potential at a point due to charge Q is:

k is a constant and k= =9× 109 Nm2C-2 . q is the point charge.
If we substitute value of k then:


Hence, Electric potential at a point x from the center of the ring is .
Net electric field at P is Ecosθ and
From the figure:






Question 64.

An electric field of magnitude 1000 NC–1 is produced between two parallel plates having a separation of 2.0 cm as shown in figure.

(a) What is the potential difference between the plate?

(b) With what minimum speed should an electron be projected from the lower plate in the direction of the field so that it may reach the upper plate?

(c) Suppose the electron is projected from the lower plate with the speed calculated in part (b). The direction of projection makes an angle of 60° with the field. Find the maximum height reached by the electron.






Answer:


Given:
Magnitude of the electric field : E = 1000 NC–1
Separation between the plates: r = 2 cm = 0.02 m
Angle made by the projection with the field : θ = 60°

Formula used:
(a)
The Potential difference is:
V = -E.r
Here, E is the electric field and r is the separation between the plates.
V = - 1000 × 0.2
∴ V = -200 = |-200| = 200 V


Hence, the potential difference between the plates is of 200 V
(b)
Charge on an electron: e = -1.6× 10-19 C
We know that
F=qE=ma
Where F is the electric force, E is the electric field, m is the mass of the body and a is the acceleration of the body.
Here, q=e of electron. And m = 9.7× 10-31 kg mass of the electron.


Using one of the equations of motion we get,
v2=u2+2as
Here, v is the final velocity of the electron= 0
Here v is zero as it we have to calculate u when it just reaches the upper plate, u is the initial velocity of the electron, s is the distance between the plates: s=r and a is the acceleration of the electrons


Hence, with a minimum speed of 2.64× 106 m/s2 should the electron be projected from the lower plate in the direction of the field for it to reach the upper plate.
(c)
As direction of projection makes an angle 60° with the electric field.
Initial velocity is resolved into its cos component which is along the direction of the electric field as shown in the figure above.
Hence, u’=ucos(60° ), u’ is the resolved initial velocity.
Using the same equation of motion used in (b) we get,
Where h is the maximum height reached by the electron.

Hence, the maximum height reached by the electron is 4.7× 10-3 m.



Question 65.

A uniform field of 2.0 NC–1 exists in space in x-direction.

(a) Taking the potential at the origin to be zero, write an expression for the potential at a general point (x, y, z).

(b) At which points, the potential is 25 V?

(c) If the potential at the origin is taken to be 100 V, what will be the expression for the potential at ta general point?

(d) What will be the potential at the origin if the potential at infinity is taken to be zero? Is it practical to choose the potential at infinity to be zero?



Answer:


Given:
(a)Magnitude of Electric field: E = 2.0 NC–1
V(0,0,0) = 0 V
(b) V= 25 V
(c) V(0,0,0) = 100 V
Formula used:
(a) Electric field exists in x-direction.
We know that, Potential difference is :

V=VB-VA=VB-0=VB

Here VB is the potential at general point (x,y,z) and VA is the potential at origin = 0.
Potential is given a
V = -E.r
Where E is the electric field and r is the position of the point in space.
In vector form:

Substituting ,

Here, y and z components are not considered as electric field is along x direction.
Hence expression of potential at a general point (x,y,z) in space is -2x V.
(b)
Here, VB is 25 V.


Hence at x= -12.5 m, potential is 25 V.
(c)
Now similar to part (a), here potential at origin is given and we need to find potential at general point.
V(0,0,0) = 100 V
Using formula for potential derived in (a) we get,





Let potential at infinity be : V = 0 and x=∞
Potential at origin is : V0
Using the formula for potential and result of (a):



Hence, potential at origin is infinite.
It is not practical to choose potential at infinity to be zero as it will make potential at origin to be infinite as we derived which will make the calculations impossible.



Question 66.

How much work has to be done in assembling three charged particles at the vertices of an equilateral triangle as shown in figure?




Answer:


Given:


Charge at 1 : q1 = 4.0 × 10-5 C
Charge at 2 : q2 = 2.0 × 10-5 C
Charge at 3 : q3 = 3.0 × 10-5 C
Let side of the equilateral triangle be L = 10 cm = 0.1 m
Formula used:
To assemble the charges at three vertices electric potential energy is required.
Thus, the work done in assembling these charges is equal to the total electric potential energy used.
W = U12 +U23 + U31
Here, W is the work done.
U12: Electric Potential energy due to charges at 1 and 2 having separation 0.1 m.
U23 : Electric Potential energy due to charges at 2 and 3 having separation 0.1 m.
U31: Electric Potential energy due to charges at 3 and 1 having separation 0.1 m.
Formula for Electric potential energy is:
Here, k is a constant and k= =9× 109 Nm2C-2 , q1 and q2 are point charges and r is the separation between them and U is the electric potential energy required to moves charges q1 and q2 apart.
Substituting,



Hence, 234 J of work is required for assembling the charges at the vertices of the equilateral triangle.



Question 67.

The kinetic energy of a charged particle decreases by 10 J as it moves from a point at potential 100V to a point at potential 200V. Find the charge on the particle.


Answer:

Given: Change in Kinetic Energy of a charged particle: ΔE = 10 J
Potential difference : dV = 200-100 =100 V
We know that, when energy of a body is changed, work is being done.
Change in kinetic energy = Work Done
Thus, Work done : W = 10 J
Also Work is given as:
W = dV× q
Here dV is the potential difference and q is the charge of the particle.
Substituting the values:

Hence the charge on the particle is 0.1 C



Question 68.

Two identical particles, each having a charge of 2.0 × 10–4 C and mass of 10g, are kept at a separation of 10 cm and then released. What would be the speeds of the particles when the separation becomes large?


Answer:


Given:
Charge on two identical particles: q1=q2= 2.0 × 10–4 C
Mass of the two particles : m1=m2= m = 10 g = 0.01 kg
Separation between the charges : r = 10 cm = 0.1 m
Formula used:
When they are released, the force of repulsion will be acting on both the particles making them drift apart thereby increasing distance between them.
Potential Energy at the start will be
where Uinitial is the initial potential energy between the charges ,k is a constant and k= =9× 109 Nm2C-2 , q1 and q2 are point charges and r is the separation between them.
Final Potential energy would be zero as the particles will gain Kinetic energy and the separation between the charges would approach infinity : Ufinal = 0
Now by conservation of energy:
Total kinetic energy is :
Where , m is the mass of the particles and V is the velocity gained by the particles.





Hence, when the separation becomes large the speed of the particles should be 600 m/s.



Question 69.

Two particles have equal masses of 5.0g each and opposite charges of +4.0 × 10–5 C and –4.0 × 10–5 C. They are released from rest with a separation of 1.0 m between them. Find the speeds of the particles when the separation is reduced to 50 cm.


Answer:


Given:
Mass of the two particles : m1=m2= m = 5.0 g = 0.005 kg
Charge on particle 1 : q1 = +4.0 × 10–5 C
Charge on particle 2 : q2 = -4.0 × 10–5 C
Separation between the charges : r = 1 m
Initial velocity: v = 0
Formula used:
Conservation of Energy is:
Uinitial + K.Einitial = Ufinal + K.Efinal
where, Uinitial and K.Einitial are the Initial potential energy and initial kinetic energy respectively .
Ufinal and K.E final are the final potential energy and final kinetic energy respectively.
Now, as initial velocity is zero : K.Einitial = 0
Final Kinetic Energy of both the particles is:
Also, Electric potential energy is given as;
Where k is a constant and k= =9× 109 Nm2C-2 , q1 and q2 are point charges and r is the separation between them.
For initial situation : r = 1 m
For final situation : r = 50 cm = 0.5 m = r/2
Substituting in the conservation formula we get,






Hence the velocity of the particles when separation is reduced to 50 cm is 53.66 m/s



Question 70.

A sample of HCl gas is placed in an electric field of 2.5 × 104 N C–1. The dipole moment of each HCl molecule is 3.4 × 10–30 Cm. Find the maximum torque that can act on a molecule.


Answer:


Given:
Electric field : E = 2.5 × 104 N C–1
Dipole moment of each HCl molecule :
P = 3.4 × 10–30 cm = 3.4 × 10–30 Cm .

Formula used:
Torque acting on a dipole is given as
Where, τ is the torque acting on the dipole, P is the dipole moment of the HCl molecules and E is the electric field.
τ = PEsinθ
For maximum torque θ = 90°
Thus sin(90) = 1


Hence a maximum of 8.5× 10-26 Nm of torque can act on a HCl molecule.



Question 71.

Two particles A and B, having opposite charges 2.0 × 10–6 C and –2.0 × 10–6 C, are placed at a separation of 1.0 cm.

(a) Write down the electric dipole moment of this pair.

(b) Calculate the electric field at a point on the axis of the dipole 1.0 m away from the center.

(c) Calculate the electric field at a point on the perpendicular bisector of the dipole and 1.0 m sway from the center.



Answer:


Given:
(a)
Charge on particle A : q1 = 2.0 × 10–6 C
Charge on particle B : q2 = -2.0 × 10–6 C
Magnitude of both the charges : q = 2.0 × 10–6 C
Separation between A and B : d = 1.0 cm = 0.01 m
(b)
Distance between center and the point on the axis of dipole: r= 1 cm= 0.01 m
(c)
Distance between center and the point on the perpendicular bisector of the dipole:
r’= 1 m
Formula used:
(a)
Electric dipole moment is given as:
Where, is the electric dipole moment, q is the magnitude of the charges at the end of the dipole and is the vector joining the two charges.


Hence, electric dipole moment between A and B is 2× 10-8Cm
(b)
Electric field at a point on the axis of the dipole is:
Here k is a constant and k= =9× 109 Nm2C-2, p is the Electric dipole moment and r is the distance between the point on the axis and it’s center.
Substituting we get,

Hence, electric field at a distance 1 cm away from the center of the dipole to the point on it’s axis is 3.6× 108 NC-1.
(c)
Electric field at a point on the perpendicular bisector of the dipole is given as:
Here, k is a constant and k= =9× 109 Nm2C-2, p is the Electric dipole moment and r’ is the distance between the point on the perpendicular bisector of the dipole and it’s center
Substituting we get:

Hence, electric field at a point on the perpendicular bisector of the dipole 1 m away from it’s center is 180 NC-1.



Question 72.

Three charges are arranged on the vertices of n equilateral triangle as shown in figure. Find the dipole moment of the combination.



Answer:


Given:
Distance between two charges :d

In the diagram, the dipole moments are extended and resolved into the resultant component .
Formula used:
The resultant is given as vector sum of
As angle between and is 30° , the value of the resultant:
As p1=p2, the resultant due to is :
We know that:
p=qd
Where p is the dipole moment, q is the magnitude of charges and d is the distance between the charges forming the dipole.
Substituting, the resultant dipole moment is:

Hence, dipole moment of the combination is qd√3



Question 73.

Find the magnitude of the electric field at the point P in the configuration shown in figure for d >> a. Take 2qa = p.



Answer:


Given:
(a)
Distance between charge and P : d
(b)
Distance between two charges +q and -q = 2a
Dipole moment : p =2qa
Formula used:
(a)
It’s a direct formula of the electric field due to a point charge:
Here, k is a constant and k= =9× 109 Nm2C-2 . q is the point charge and d is the distance between the charge and the point P
(b)
For the second configuration too we have a direct formula.
Opposite charges at both ends separated by a distance forms a dipole and electric field at a point on the perpendicular bisector of the dipole is given as:
Here, p is the dipole moment and is : p= 2qa and k is a constant and k= =9× 109 Nm2C-2 , d is the distance between Point P and the center of the dipole.
(c)
Figure (c) is made up of combination of system in (a) and (b)

Point P is influenced by two dipoles.
1.-q and +q (A and B)
2. +q and +q (B and C)
1. Is the result of (b) = E2
2. Is the result of (a) = E1
Hence net electric field at P in the configuration (c) would be resultant of E1 and E2






Hence, results for (a), (b) and (c) are ,, and respectively.



Question 74.

Two particles, carrying –q and +q and having equal masses m each, are fixed at the ends of a light rod of length a to form a dipole. The rod is clamped at an end and is placed in a uniform electric field E with the axis of the dipole along the electric field. The rod is slightly tilted and then released. Neglecting gravity find the time period of small oscillations.


Answer:


Given:
Magnitude of charge on both the particles: q
Mass of both the particles: m
Electric field : E
The axis of the dipole is along the electric field.
Length of the dipole: l=a

Formula used:
We know that, a charge placed in an electric field E will experience electric force.
When a dipole with charge of magnitude q at its ends is suspended under a magnetic field and having one end clamped (-q), the free end (+q) will experience an electric force and it will start oscillating.
Electric force is given as:
F=qE
and
F = ma
∴ qE = ma
Where a is the acceleration of the particle, q is the charge on the particle, m is the mass of the particle and E is the electric field.
Also, time period for Simple pendulum is:
Here, g=a(acceleration) as we are neglecting gravity.
Substituting we get,

Hence, time period of the small oscillations is .



Question 75.

Assume that each atom in a copper wire contributes one free electron. Estimate the number of free electrons in a copper wire having a mass of 6.4 g (take the atomic weight of copper to be 64 g mol–1).


Answer:


Given:
Atomic weight of copper = 64 g mol-1
Mass of copper wire : m = 6.4 g
Formula used:
We know that,
Number of moles in 64 g of copper = 1
Thus, number of moles in 6.4 g of copper = 0.1
Also,
Number of atoms in one mole=NA
Here, NA is the Avogadro Number : NA = 6.023× 1023 atoms
∴No.of atoms in 0.1 mole of copper=0.1× 6.023× 1023
∴No.of atoms in 0.1 mole of copper=0.1× 6.023× 1022
It is given that each atom contributes one free electron,
Therefore in 0.1 mole of copper will contribute
Hence, there will be free electron in 6.4 g of copper.