Electric Charges and Fields

Class 12 Physics Chapter 1 Important Questions Electric Charges and Fields

Very Short Answer

Question 1.
What is the value of the angle between the vectors \(\vec{p}\) and \(\vec{E}\) for which the potential energy of an electric dipole of dipole moment \(\vec{p}\), kept in an external electric field \(\vec{E}\), has maximum value.
Answer:
P.E. = –pEcos θ
P.E. is maximum when cos θ = – 1, i.e.
θ = 180°

Question 2.
Define electric field intensity at a point.
Answer:
Electric field intensity at a point is defined as the force experienced by a unit test charge placed at that point. Mathematically
we have

Question 3.
Two equal point charges separated by 1 m distance experience force of 8 N. What will be the force experienced by them, if they are held in water, at the same distance? (Given: K_water = 80)
Answer:
The force in water is given by
F_w = \(\frac{F_{\text {air }}}{K}=\frac{8}{80}\) = 0.1 N

Question 3 a.
Which orientation of an electric dipole in a uniform electric field would correspond to stable equilibrium ? (All India 2008)
Answer:
When dipole moment vector is parallel to electric field vector
\(\overrightarrow{\mathrm{P}} \| \overrightarrow{\mathrm{E}}\)

Question 4.
If the radius of the Gaussian surface enclosing a charge is halved, how does the electric flux through the Gaussian surface change ?
Answer:
Electic flux ϕ_E is given by
\(\phi_{\mathrm{E}}=\oint \overrightarrow{\mathrm{E}} \cdot d \vec{s}=\frac{\mathrm{Q}}{\varepsilon_{0}}\)
…. where [Q is total charge inside the closed surface
∴ On changing the radius of sphere, the electric flux through the Gaussian surface remains same.

Question 5.
Define the term electric dipole moment of a dipole. State its S.I. unit
Answer:
τ = OE sin θ
If E = 1 unit, θ = 90°, then τ = P
Dipole moment may be defined as the torque acting on an electric dipole, placed perpendicular to a uniform electric dipole, placed perpendicular to a uniform electric field of unit strength.
or Strength of electric dipole is called dipole moment.
\(|\overrightarrow{\mathrm{P}}|=q|2 a|\)
∴ SI unit is Cm.

Question 6.
In which orientation, a dipole placed in a uniform electric field is in

• stable,
• unstable equilibrium ? (Delhi 2010)

Answer:

• For stable equilibrium, a dipole is placed parallel to the electric field.
• For unstable equilibrium, a dipole is placed antiparallel to the electric field.

Question 7.
Figure shows three point charges, +2q, -q and + 3q. Two charges +2q and -q are enclosed within a surface ‘S’. What is the electric flux due to this configuration through the surface ‘S’ (Delhi 2010)

Answer:
\(\text { Electric flux }=\oint_{\mathrm{S}} \overrightarrow{\mathrm{E}} \cdot \overrightarrow{d \mathrm{S}}\)

Question 8.
A charge ‘q’ is placed at the center of a cube of side l. What is the electric flux passing through each face of the cube?
Answer:
Φ = q/6ε₀

Question 9.
Why do the electric field lines not form closed loops?
Answer:
It is due to the conservative nature of the electric field.

Question 10.
Two equal balls having equal positive charge ‘q’ coulomb are suspended by two insulating strings of equal length. What would be the effect on the force when a plastic sheet is inserted between the two?
Answer:
It decreases because force ∝= \(\frac{1}{k}\) and k > 1.

Question 11.
What is the electric flux through a cube of side l cm which encloses an electric dipole?
Answer:
Zero

Question 12.
Why are electric field lines perpendicular at a point on an equipotential surface of a conductor?
Answer:
So that no net force acts on the charge at the equipotential surface and it remains stationary.

Question 13.
What is the amount of work done in moving a point charge Q. around a circular arc of radius ‘r’ at the centre of which another point charge ‘q’ is located?
Answer:
Zero.

Question 14.
How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased?
Answer:
No change, as flux does not depend upon the size of the Gaussian surface.

Question 15.
Name the physical quantity whose S.I. unit is JC^-1. Is it a scalar or a vector quantity? (All India 2010)
Answer:

• Physical quantity whose S.I. unit is JC^-1 is Electric potential.
• It is a Scalar quantity.

Question 16.
Define electric dipole moment. Write its S.I. unit. (All India 2011)
Answer:
Electric dipole moment of an electric dipole is defined as the product of the magnitude of either charge and dipole length.

S.I. unit of dipole (\(\vec{p}\)) is coulomb metre (Cm).

Question 17.
Why should electrostatic field be zero inside a conductor? (Delhi 2012)
Answer:
Electrostatic field inside a conductor should be zero because of the absence of charge. As in a static condition, charge remains only on the surface.

Question 18.
Why must electrostatic field be normal to the surface at every point of a charged conductor? (Delhi 2012)
Answer:
So that tangent on charged conductor gives the direction of the electric field at that point.

Question 19.
A charge ‘q’ is placed at the centre of a cube of side l. What is the electric flux passing through each face of the cube? (All India 2012)
Answer:
Electric flux through each phase of the cube

Question 20.
A charge ‘q’ is placed at the centre of a cube of side l. What is the electric flux passing through two opposite faces of the cube? (All India)
Answer:
\(\phi_{\mathrm{E}}=\frac{q}{3 \varepsilon_{0}}\)

Question 21.
Depict the direction of the magnetic field lines due to a circular current carrying loop. (Comptt. Delhi 2012)
Answer:
Direction of the magnetic field lines is given by right hand thumb rule.

Question 22.
What is the direction of the electric field at the surface of a charged conductor having charge density σ < 0? (Comptt. Delhi 2012)
Answer:
The direction of electric field is normal and inward to the surface.

Question 23.
Why do the electric field lines not form closed loops?
Answer:
Electric field lines do not form closed loops because the direction of an electric field is from positive to negative charge. So one can regard a line of force starting from a positive charge and ending on a negative charge. This indicates that electric field lines do not form closed loops.

Question 24.
Is the electric field due to a charge configuration with total charge zero, necessarily zero? Justify. (Comptt. All India 2012)
Answer:
No, it is not necessarily zero. If the electric field due to a charge configuration with total charge is zero because the electric field due to an electric dipole is non-zero.

Question 25.
Two charges of magnitudes – 2Q and + Q are located at points (a, 0) and (4a,0) respectively. What is the electric flux due to these charges through a sphere of radius ‘3a’ with its centre at the origin? (All India 2013)
Answer:

Question 26.
Two charges of magnitudes -3Q and + 2Q are located at points (a, 0) and (4a, 0) respectively. What is the electric flux due to these charges through a sphere of radius ‘5a’ with its centre at the origin?
Answer:

Question 27.
Write the expression for the work done on an electric dipole of dipole moment p in turning it from its position of stable equilibrium to a position of unstable equilibrium in a uniform electric
field E. (Comptt. Delhi 2013)
Answer:
Torque, acting on the dipole is, τ = pE sin θ

Question 28.
Why do the electrostatic field lines not form closed loops? (All India 2014)
Answer:
Electric field lines do not form closed loops because the direction of an electric field is from positive to negative charge. So one can regard a line of force starting from a positive charge and ending on a negative charge. This indicates that electric field . lines do not form closed loops.

Question 29.
Draw the pattern of electric field lines, when a point charge –Q is kept near an uncharged conducting plate.
Answer:
The pattern is as shown.

Question 30.
Draw a plot showing the variation of the electric field with distance from the center of a solid conducting sphere of radius R, having a charge + Q on its surface.
Answer:
The plot is as shown.

Question 31.
Draw a graph to show the variation of E with perpendicular distance r from the line of charge (CBSE Delhi 2018)
A
Answer:
E = \(\frac{\lambda}{2 \pi \varepsilon_{0} r}\)
E ∝ \(\frac{1}{r}\)

The graph between E and r is as shown.

Question 32.
Define the term ‘electric flux’. Write its S.I. unit.
Answer:
Electric flux: It is the measure of the number of electric field lines crossing a given area normally.

Mathematically the electric flux passing through an area \(\vec{dS}\) is given by
dΦ = \(\vec{E}\) . \(\vec{dS}\)

SI unit of electric flux is Nm²C^-1 or Vm.

Question 33.
Why can the interior of a conductor have no excess charge in the static situation? (CBSE Ai 2019)
Answer:
Since the electric field inside the conductor is zero, electric flux through the closed surface is also zero. Hence by Gauss’s law, there is no net charge enclosed by the closed surface.

Question 34.
Two field lines never cross each other. Why?
Answer:
It is because at the point of intersection two perpendiculars can be drawn. Thus there will be two directions of the electric field at that point which is not possible.

Question 35.
In an electric field, an electron is kept freely. If the electron is replaced by a proton, what will be the force experienced by the proton?
Answer:
The magnitude of force will be the same but the direction will be reversed.

Question 36.
Why do the electric field lines never cross each other? (All India)
Answer:
The electric lines of force give the direction of the electric field. In case, two lines of force intersect, there will be two directions of the electric field at the point of intersection, which is not possible.

Question 37.

distance ‘d’ apart as shown in the figure. The electric field intensity is zero at a point ’P’ on the line joining them as shown. Write two conclusions that you can draw from this. (Comptt. Delhi 2014)
Answer:

• Two point charges ‘ q₁‘ and ‘ q₂ should be of opposite nature.
• Magnitude of charge ql must be greater than that of charge q₂.

Question 38. What is the electric flux through a cube of side 1 cm which encloses an electric dipole? (Delhi 2015)
Answer:
Zero because the net charge of an electric dipole (+ q and – q) is zero.

Question 39.
Why are electric field lines perpendicular at a point on an equipotential surface of a conductor? (Comptt. All India 2015)
Answer:
If the electric field lines were not normal to the equipotential surface, it would have a non-zero component along the surface. To move a unit test charge against the direction of the component of the field, work would have to be done which means this surface cannot be equipotential surface.
Hence, electric field lines are perpendicular at a point on an equipotential surface of a conductor.

Question 40.

Is the potential difference V_A – V_B positive, negative or zero? (Delhi 2016)
Answer:
The potential difference is positive.

Question 41.
How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased? (Delhi 2016)
Answer:
The electric flux due to a point charge enclosed by a spherical gaussian surface remains ‘unaffected’ when its radius is increased.

Question 42.
Show on a plot the nature of variation of the

• Electric field (E) and
• potential (V), of a (small) electric dipole with the distance (r) of the field point from the centre of the dipole. (Comptt. Outside Delhi 2016)

Answer:

Question 43.
Does the charge given to a metallic sphere depend on whether it is hollow or solid? Give reason for your answer. (Delhi 2017)
Answer:
No, it does not, because the charge resides only on the surface of the conductor.

Question 44.
Draw a plot showing variation of electric field with distance from the centre of a solid conducting sphere of radius R, having a charge of +Q on its surface. (Comptt. Delhi 2017)
Answer:
Plot between E and r

Question 45.
A point charge +Q is placed in the vicinity of a conducting surface. Draw the electric field lines between the surface and the charge.
(Comptt. Outside Delhi 2017)
Answer:

Question 46.
Derive an expression for the torque experienced by an electric dipole kept in a uniform electric field. (Delhi 2017)
Answer:
Consider an electric dipole consisting of charges + q and – q and of length 2a placed in a uniform electric field \(\overrightarrow{\mathrm{E}}\) making an angle θ with it. It has a dipole moment of magnitude,

Hence the net translating force on a dipole in a uniform electric field is zero. But the two equal and opposite forces act at different points of the dipole. They form a couple which exerts a torque.
Torque = Either force × Perpendicular distance between the two forces
x = qE × 2a sin θ
X = pE sin θ [ ∵ p = q × 2a; p is dipole moment]
As the direction of torque \(\vec{\tau}\) is perpendicular to \(\vec{p}\) and \(\vec{E}\), so we can write \(\vec{\tau}=\vec{p} \times \overrightarrow{\mathrm{E}}\)

Question 47.
Define electric flux. Write its S.I. unit.
A charge q is enclosed by a spherical surface of radius R. If the radius is reduced to half, how would the electric flux through the surface change? (All India 2009)
Answer:
Electric flux over an area in an electric field is the total number of lines of force passing through the area. It is represented by ϕ . It is a scalar quantity. Its S.I unit is Nm² C^-1 or Vm.

Electric flux ϕ by q_enclosed
Hence the electric flux through the surface of sphere remains same.

Question 48.
A spherical conducting shell of inner radius rx and outer radius r2 has a charge ‘Q’. A charge ‘q’ is placed at the centre of the shell.
(a) What is the surface charge density on the
(i) inner surface,
(ii) outer surface of the shell?
(b) Write the expression for the electric field at a point x > r₂ from the centre of the shell.
Answer:
(a) Surface charge density on the :

(b) Electric field at a point x > r₂ from the centre of the shell will be E = \(\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q+Q}{x^{2}}\right)\)

Question 49.
Show that the electric field at the surface of a charged conductor is given by \(\overrightarrow{\mathrm{E}}=\frac{\sigma}{\varepsilon_{0}} \hat{n}\), where σ is the surface charge density and h is a unit vector normal to the surface in the outward direction. (All India 2010)
Answer:
Electric field at a point on the surface of charged conductor, E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{R^{2}}\)
For simplicity we consider charged conductor as a sphere of radius ‘R’. If ‘σ’ is in surface charge density, then

…where [ \(\hat{\boldsymbol{n}}\) is a unit vector normal to the surface in the outward direction]

Question 50.
Consider the situation shown in the figure given below. What are the signs of q₁ and q₂?
Answer:

q₁ is negative and q₂ is positive.

Question 51.
In the figure given below, at which point electric field is maximum?
Answer:

The electric field is maximum at point C.

Question 52.
An uncharged insulated conductor A is brought near a charged insulated conductor B. What happens to the charge and potential of B?
Answer:
On bringing uncharged conductor A near a charged conductor B, charges are induced on A as shown in the figure below. As a result, the potential of conductor B is slightly lowered but the charge on it remains unchanged.

Question 53.
In a medium the force of attraction between two point electric charges, distance ‘d’ apart, is F. What distance apart should these be kept in the same medium so that the force between them becomes 3F?
Answer:
Let the new distance be ‘d’, since F ∝\(\frac{1}{r^{2}}\) ,
therefore \(\frac{F}{3 F}=\frac{r^{2}}{d^{2}}=\frac{1}{3} \Rightarrow r=\frac{d}{\sqrt{3}}\)

Question 54.
Find the value of an electric field that would completely balance the weight of an electron.
Answer:
mg = eE ⇒ E = \(\frac{m g}{e}\)
= \(\frac{9.1 \times 10^{-31} \times 9.8}{1.6 \times 10^{-19}}=5.57 \times 10^{-11} \mathrm{Vm}^{-1}\)

Question 55.
Two charges, one +5 µC, and the other -5 µC are placed 1 mm apart. Calculate the electric dipole moment of the system.
Answer:
p = q × 2a = 5 × 10 ^-6 × 10^-3 = 5 × 10^-9 Cm

Question 56.
Two-point charges of+3 µC each are 100 cm apart. At what point on the line joining the charges will the electric intensity be zero?
Answer:
At the mid-point of the line joining the two point charges.

Question 57.
Four charges of came magnitude and same sign are placed at the corners of a square, of each side 0.1 m. What is electric field intensity at the center of the square?
Answer:
Zero.

Question 58.
Why should the electrostatic field be zero ‘ inside a conductor?
Answer:
Because it does not contain any charge.

Question 59.
A metallic spherical shell has an inner radius R₁ and outer radius R₂. A charge Q is placed at the center of the shell. What will be the surface charge density on the (i) inner surface, and (ii) the outer surface of the shell?
Answer:
On inner surface

On the outer surface,

Question 60.
An arbitrary surface encloses a dipole. What is the electric flux through this surface?
Answer:
Zero.

Question 61.
A thin straight infinitely long conducting wire having charge density X is enclosed by a cylindrical surface of radius r and length l, its axis coinciding with the length of the wire. Find the expression for the electric flux through the surface of the cylinder. (All India 2011)
Answer:
Since the field is everywhere radial, flux through the two ends of the cylindrical Gaussian surface is zero. At the cylindrical part of the surface, E is normal to the surface at every point, and its magnitude is constant, since it depends only on r. The surface area of the curved part is 2πrl, where l is the length of the cylinder.

Flux through the Gaussian surface = Flux through the curved cylindrical part of the surface is zero. At the cylindrical part of the surface, E is normal to the surface at every point, and its magnitude is constant, since at every point, and its magnitude is constant, since it depends only on r. The surface area of the cylinder.

Flux through the Gaussian surface = Flux through the curved cylindrical part of the surface
= E × 2πrl

(a) Electric field due to an infinitely long thin straight wire is radial.
(b) The Gaussian surface for a long thin wire of uniform linear charge density
The surface includes charge equal to λl.
Gauss’s law then gives

Question 62.
Plot a graph showing the variation of coulomb force (F) versus \(\left(\frac{1}{r^{2}}\right)\), where r is the distance between the two charges of each pair of charges : (1µC, 2µC) and (2µC, – 3µC). Interpret the graphs obtained.
Answer:

Here positive slope depicts that force is repulsive in nature and negative slope depicts that the force is attractive in nature.

Question 63.
A hollow cylindrical box of length 1m and area of cross-section 25 cm² is placed in a three dimensional coordinate system as shown in the figure. The electric field in the region is given by \(\overrightarrow{\mathbf{E}}=50 x \hat{i}\) where E is in NC^-1 and x is in metres. Find

• Net flux through the cylinder.
• Charge enclosed by the cylinder.

Answer:

Question 64.
Given a uniform electric field , find the flux of this field through a square of 10 cm on a side whose plane is parallel to the y-z plane. What would be the flux through the same square if the plane makes a 30° angle with the x-axis? (Delhi 2014)
Answer:
Given : \(\overrightarrow{\mathrm{E}}=5 \times 10^{3} \hat{i} \mathrm{N} / \mathrm{C}\)
A = 10 × 10 × 10^-4m²,
Flux (ϕ) = EA cos θ
(i) For first case, θ = 0, cos 0 = 1
∴ Flux = (5 × 10³) × (10 × 10 × 10^-4)
(ii) Angle of square plane with x-axis = 30°
Hence the 0 will be 90° – 30° = 60°
EA cos θ = (5 × 10³) × (10 × 10 × 10^-4) × cos 60
= 50 × \(\frac{1}{2}\)
= 25 Nm²C^-1

Question 65.
Given a uniform electric field \(\overrightarrow{\mathbf{E}}\) = 2 × 10³ \(\hat{i}\) N/ C, find the flux of this field through a square of side 20 cm, whose plane is parallel to the y-z plane. What would be the flux through the same square, if the plane makes an angle of 30° with the x-axis? (Delhi 2014)
Answer:
Given : \(\overrightarrow{\mathrm{E}}=5 \times 10^{3} \hat{i} \mathrm{N} / \mathrm{C}\)
A = 10 × 10 × 10^-4m²,
Flux (ϕ) = EA cos θ
(i) For first case, θ = 0, cos 0 = 1
∴ Flux = (5 × 10³) × (10 × 10 × 10^-4)
(ii) Angle of square plane with x-axis = 30°
Hence the 0 will be 90° – 30° = 60°
EA cos θ = (5 × 10³) × (10 × 10 × 10^-4) × cos 60
= 50 × \(\frac{1}{2}\)
= 25 Nm²C^-1
Hint : (i) 80 Nm²C^-1
(ii) 40 Nm²C³

Question 66.
Given a uniform electric field \(\overrightarrow{\mathrm{E}}=4 \times 10^{3} \hat{i}\) N/C. Find the flux of this field through a square of 5 cm on a side whose plane is parallel to the Y-Z plane. What would be the flux through the same square if the plane makes a 30° angle with the x-axis? (Delhi 2014)
Answer:
Given : \(\overrightarrow{\mathrm{E}}=5 \times 10^{3} \hat{i} \mathrm{N} / \mathrm{C}\)
A = 10 × 10 × 10^-4m²,
Flux (ϕ) = EA cos θ
(i) For first case, θ = 0, cos 0 = 1
∴ Flux = (5 × 10³) × (10 × 10 × 10^-4)
(ii) Angle of square plane with x-axis = 30°
Hence the 0 will be 90° – 30° = 60°
EA cos θ = (5 × 10³) × (10 × 10 × 10^-4) × cos 60
= 50 × \(\frac{1}{2}\)
= 25 Nm²C^-1
Hint:
(i) 10 Nm²C^-1
(ii) 5 Nm²C^-1

Question 67.
A small metal sphere carrying charge +Q is located at the centre of a spherical cavity in a large uncharged metallic spherical shell. Write the charges on the inner and outer surfaces of the shell. Write the expression for the electric field at the point P₁ (Comptt. Delhi)

Answer:

1. Charge on inner surface : – Q
2. Charge on outer surface : + Q
3. Electric field at point P₁ (E) = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r_{1}^{2}}\)

Question 68.
An electric dipole is placed in a uniform electric field \(\overrightarrow{\mathbf{E}}\) with its dipole moment \(\vec{p}\) parallel to the field. Find
(i) the work done in turning the dipole till its dipole moment points in the direction
opposite to \(\overrightarrow{\mathbf{E}}\) .
(ii) the orientation of the dipole for which the torque acting on it becomes maximum.
Answer:

(ii) τ = \(\vec{p} \times \overrightarrow{\mathrm{E}}\) = pE sin θ
For θ = \(\frac{\pi}{2}\), sin θ = 1 and τ is maxximum

Question 69.
A sphere S₁ of radius r₁ encloses a net charge Q. If there is another concentric sphere S₂ of radius r₂ (r₂ > r,) enclosing charge 2Q, find the ratio of the electric flux through S₁ and S₂. How will the electric flux through sphere S₁ change if a medium of dielectric constant K is introduced in the space inside S₂ in place of air? (Comptt. All India 2014)

Answer:

Therefore, there will be no change in the flux through S₁ on introducing dielectric medium inside the sphere S₂.

Question 70.
Define the term ‘electric flux’. Write its S.I. units. What is the flux due to electric field \(\overrightarrow{\mathbf{E}}=3 \times 10^{3} \hat{i}\) N/C through a square of side 10 cm, when it is held normal to if? (Comptt. All India 2015)
Answer:

Electric flux over an area in an electric field is the total number of lines of force passing through the area. It is represented by ϕ . It is a scalar quantity. Its S.I unit is Nm² C^-1 or Vm.

Electric flux ϕ by q_enclosed
Hence the electric flux through the surface of sphere remains same.

Short Answer Type

Question 1.
(a) Electric field inside a conductor is zero. Explain.
(b) The electric field due to a point charge at any point near it is given as
E  =

what is the physical significance of this limit?
Answer:
(a) By Gauss theorem \(\phi \vec{E} \cdot \overrightarrow{d S}=\frac{q}{\varepsilon_{0}}\). Since there is no charge inside a conductor therefore in accordance with the above equation the electric field inside the conductor is zero.
(b) It indicates that the test charge should be infinitesimally small so that it may not disturb the electric field of the source charge.

Question 2.
Define the electric line of force and give its two important properties.
Answer:
It is a line straight or curved, a tangent to which at any point gives the direction of the electric field at that point.
(a) No two field lines can cross, because at the point of intersection two tangents can be drawn giving two directions of the electric field which is not possible.
(b) The field lines are always perpendicular to the surface of a charged conductor.

Question 3.
Draw electric field lines due to (i) two similar charges, (ii) two opposite charges, separated by a small distance.
Answer:
(a) The diagram is as shown.

(b) The diagram is as shown.

Question 4.
An electric dipole is free to move in a uniform electric field. Explain what is the force and torque acting on it when it is placed
(i) parallel to the field
Answer:
When an electric dipole is placed parallel to a uniform electric field, net force, as well as net torque acting on the dipole, is zero and, thus, the dipole remains in equilibrium.

(ii) perpendicular to the field
Answer:
When the dipole is placed perpendicular to the field, two forces acting on the dipole form a couple, and hence a torque acts on it which aligns its dipole along the direction of the electric field.

Question 5.
A small metal sphere carrying charge +Q. is located at the center of a spherical cavity in a large uncharged metallic spherical shell. Write the charges on the inner and outer surfaces of the shell. Write the expression for the electric field at the point P₁
Answer:

Charge on inner surface – Q.
Charge on outer surface + Q,
Electric field at point P = E = k\(\frac{Q}{r_{1}^{2}}\)

Question 6.
Two-point charges q and –2q are kept ‘d’ distance apart. Find the location of the point relative to charge ‘q’ at which potential due to this system of charges is zero.
Answer:
Let the potential be zero at point P at a distance x from charge q as shown

Now potential at point P is
V = \(\frac{k q}{x}+\frac{k(-2 q)}{d+x}\) = 0

Solving for x we have
x = d

Question 7.
A thin conducting spherical shell of radius R has charge Q spread uniformly over its surface. Using Gauss’s law, derive an expression for an electric field at a point outside the shell.
Draw a graph of electric field E(r) with distance r from the centre of the shell for 0 ≤ r ≤ ∞
Answer:
Electric field at a point outside the shell :
(a) (i) To find out electric field at a point outside a spherical charged shell we imagine a symmetrical Gaussian surface in such a way that the point lies on it.

Graph of electric field E(r) :

Question 8.
State Gauss’ law in electrostatics. Using this law derive an expression for the electric field due to a uniformly changed infinite plane sheet.
Answer:
Gauss’ Law states that “the total flux through a closed surface is \(\frac{1}{\varepsilon_{0}}\) times the net charge enclosed by

Let σ be the surface charge density (charge per unit area) of the given sheet and let P be a point at distance r from the sheet where we have to find \(\overrightarrow{\mathrm{E}}\)

Choosing point P’, symmetrical with P on the other side of the sheet, let us draw a Gaussian cylindrical surface cutting through the sheet as shown in the diagram. As at the cylindrical part of the Gaussian surface, \(\overrightarrow{\mathrm{E}}\) and \(d \overrightarrow{\mathrm{S}}\) are at a right angle, the only surfaces having \(\overrightarrow{\mathrm{E}}\) and \(d \overrightarrow{\mathrm{S}}\) parallel are the plane ends

…[As E is outgoing from both plane ends, the flux is positive.
This is the total flux through the Gaussian surface.

This value is independent of r. Hence, the electric field intensity is same for all points near the charged sheet. This is called uniform electric field intensity.

Question 9.
State ‘Gauss law’ in electrostatics. Use this law to derive an expression for the electric field due to an infinitely long straight wire of linear charge density λ cm^-1
Answer:
Gauss’s law in electrostatics : It states that “the total electric flux over the surface S in vaccum is \(\frac{1}{\varepsilon_{0}}\) times the total charge (q).”

Electric field due to an infinitely long straight wire : Consider an infinitely long straight line charge having linear charge density X to determine its electric field at distance r. Consider a cylindrical Gaussian surface of radius r and length l coaxial with the charge. By symmetry, the electric field E has same magnitude at each point of the curved surface S₁ and is directed radially outward.
Total flux through the cylindrical surface,

Question 10.
A positive point charge (+ q) is kept in the vicinity of an uncharged conducting plate. Sketch electric field lines originating from the point on to the surface of the plate.
Derive the expression for the electric field at the surface of a charged conductor. (All India) Answer: Representation of electric field, (due to a positive charge)
Answer:
Representation of electric field. (due to a positive charge)

Question 11.
Use Gauss’s law to derive the expression for the electric field between two uniformly charged large parallel sheets with surface charge densities a and -a respectively. (All India)
Answer:

Gauss’ Law states that “the total flux through a closed surface is \(\frac{1}{\varepsilon_{0}}\) times the net charge enclosed by

Let σ be the surface charge density (charge per unit area) of the given sheet and let P be a point at distance r from the sheet where we have to find \(\overrightarrow{\mathrm{E}}\)

Choosing point P’, symmetrical with P on the other side of the sheet, let us draw a Gaussian cylindrical surface cutting through the sheet as shown in the diagram. As at the cylindrical part of the Gaussian surface, \(\overrightarrow{\mathrm{E}}\) and \(d \overrightarrow{\mathrm{S}}\) are at a right angle, the only surfaces having \(\overrightarrow{\mathrm{E}}\) and \(d \overrightarrow{\mathrm{S}}\) parallel are the plane ends

…[As E is outgoing from both plane ends, the flux is positive.
This is the total flux through the Gaussian surface.

This value is independent of r. Hence, the electric field intensity is same for all points near the charged sheet. This is called uniform electric field intensity.

Question 12.
State Gauss’s law.
A thin straight infinitely long conducting wire of linear charge density ‘X’ is enclosed by a cy¬lindrical surface of radius V and length ‘l’—its axis coinciding with the length of the wire. Obtain the expression for the electric field, indi¬cating its direction, at a point on the surface of the cylinder. (Comptt. Delhi 2012)
Answer:
Gauss’s law. Gauss law states that “Total flux (electric flux) over the closed surfaces in vacuum is He0 times the total charge (Q) contained inside S.”

Since the field is everywhere radial, flux through the two ends of the cylindrical Gaussian surface is zero. At the cylindrical part of the surface, E is normal to the surface at every point, and its magnitude is constant, since it depends only on r. The surface area of the curved part is 2πrl, where l is the length of the cylinder.

Flux through the Gaussian surface = Flux through the curved cylindrical part of the surface is zero. At the cylindrical part of the surface, E is normal to the surface at every point, and its magnitude is constant, since at every point, and its magnitude is constant, since it depends only on r. The surface area of the cylinder.

Flux through the Gaussian surface = Flux through the curved cylindrical part of the surface
= E × 2πrl

(a) Electric field due to an infinitely long thin straight wire is radial.
(b) The Gaussian surface for a long thin wire of uniform linear charge density
The surface includes charge equal to λl.
Gauss’s law then gives

Question 13.
(a) Define electric flux. Write its S.I. units.
(b) Consider a uniform electric field
\(\overrightarrow{\mathrm{E}}\) = 3 × 10³ \(\hat{\mathbf{i}}\) N/C. Calculate the flux of this field through a square surface of area 10 cm² when
(i) its plane is parallel to the y-z plane, and
(ii) the normal to its plane makes a 60° angle with the x-axis. (Comptt. Delhi 2013)
Answer:
(a)
Electric flux over an area in an electric field is the total number of lines of force passing through the area. It is represented by ϕ . It is a scalar quantity. Its S.I unit is Nm² C^-1 or Vm.

Electric flux ϕ by q_enclosed
Hence the electric flux through the surface of sphere remains same.

Question 14.
Two charged spherical conductors of radii R₁ and R₂ when connected by a conducting wire acquire charges q₁ and q₂ respectively. Find the ratio of their surface charge densities in terms of their radii. (Delhi 2014)
Answer:
Two charged spherical conductors of radii R₁ and R₂ when connected by a conducting wire acquire charges q₁ and q₂ respectively. Thus these two conductors have a common potential V.
Two charged spherical conductors of radii R1 and R2 when connected by a conduction wire acquire charges q1 and q2 respectively. Thus these two conductors have a common potential V

Question 15.
Two point charges + q and -2q are placed at the vertices ‘B’ and ‘C’ of an equilateral triangle ABC of side as given in the figure. Obtain the expression for (i) the magnitude and (ii) the direction of the resultant electric field at the vertex A due to these two charges.

(Comptt All India)
Answer:

Question 16.
Two point charges + 3q and – 4q are placed at the vertices ‘B’ and ‘C’ of an equilateral triangle ABC of side ‘a’ as given in the figure. Obtain the expression for

(i) the magnitude and
(ii) the direction of the resultant electric field at the vertex A due to these two charges. (Comptt. All India 2014)
Answer:

Question 17.
An electric dipole of dipole moment \(\overrightarrow { p } \) is placed in a uniform electric field \(\overrightarrow { E } \)?. Obtain the expression for the torque \(\overrightarrow { \tau } \)experienced by the dipole. Identify two pairs of perpendicular vectors in the expression. (Comptt. Delhi 2015)
Answer:
(i)
(a) Torque on electric dipole. Consider an electric dipole consisting of two equal and opposite point charges separated by a small distance 2a having dipole moment

So net force on the dipole is zero
Since is uniform, hence the dipole does not undergo any translatory motion.

These forces being equal, unlike and parallel, from a couple, which rotates the dipole in clock-wise direction
∴ Magnitude of torque = Force × arm of couple

[The direction of \(\vec{\tau} \) is given by right hand screw rule and is normal to \(\vec{p} \)  ] and \(\vec{E} \)
Special cases
(i) when θ = 0 then τ = PE sin θ = 0
∴ Torque is zero and the dipole is in stable equilibrium
(ii) When θ = 90 then τ = PE sin 90 = PE
∴ The Torque is maximum
(b) Ratio of flux


∴ Electric flux through the sphere S₁ decreases with the introduction of dielectric inside it.

(ii) Two pairs of perpendicular vectors are,
(a) \(\overrightarrow { \tau } \) is perpendicular to \(\overrightarrow { p } \)
(b) \(\overrightarrow { \tau } \)is perpendicular to \(\overrightarrow { E } \)

Question 18.
(a) Two spherical conductors of radii Ra and R2 (R₂ > R₁) are charged. If they are connected by a conducting wire, find out the ratio of the surface charge densities on them.
(b) A steady current flows in a metallic conductor of non-uniform cross-section. Which of these quantities is constant along the conductor : current, current density, electric field, drift speed? (Comptt. Delhi 2015)
Answer:
(a) When two charged spherical conductors of Radii R₁ and R₂ respectively (R₂ > R₁) are connected by a conducting wire, we know that the common potential (V) is given by,

Question 19.
A charge is distributed uniformly over a ring of radius ‘a’. Obtain an expression for the electric intensity E at a point on the axis of the ring. Hence show that for points at large distances from the ring, it behaves like a point charge. (Delhi 2016)
Answer:
Electric Intensity on the axis of a ring:

This is the Electric Field due to a point charge at distance x.

Question 20.
Two thin concentric and coplanar spherical shells, of radii a and b (b > a) carry charges, q and Q, respectively. Find the magnitude of the electric field, at a point distant x, from their common centre for
(i) 0 < x < a
(ii) a ≤ x < b
(iii) b ≤ x < ∞ (Comptt. Delhi 2016) Answer: Magnitude of Electric field : Two thin concentric and coplanar spherical shells of radii ‘a’ and ‘b’ (b > a) carry charges ‘q’ and ‘Q’ respectively.
(i) For 0 < x < a
Point lies inside both the spherical shells.
Hence, E(x) = 0

(ii) For a ≤ x < b
Point is outside the spherical shell of radius ‘a’ but inside the spherical shell of radius ‘b’.
∴ E(x) = \(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{x^{2}}\)
(iii) For b ≤
x < ∞
Point is outside of both the spherical shells. Total effective charge at the centre equals (Q + q).
E(x) = \(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{(q+Q)}{x^{2}}\)

Question 21.
A charge +Q, is uniformly distributed within a sphere of radius R. Find the electric field, due to this charge distribution, at a point distant r from the centre of the sphere where :
(i) 0 < r < R and
(ii) r > R (Comptt. Outside Delhi )
Answer:
We have

Question 22.
(i) Derive the expression for electric field at a point on the equatorial line of an electric dipole.
(ii) Depict the orientation of the dipole in
(a) stable,
(b) unstable equilibrium in a uniform electric field. (Delhi 2017)
Answer:
(i)

Electric dipole moment: It is the product of the magnitude of either charge and distance between them.

It is a vector quantity whose direction is from negative to positive charge.

Electric field intensity at P due to +q charge is

Electric field intensity at P due to -q charge is,

(ii) (a) For stable equilibrium, the angle between p and E is 0°,

(b) For unstable equilibrium, the angle between p and E is 180°,

Question 23.
(i) Obtain the expression for the torque \(\vec{\tau} \) experienced by an electric dipole of dipole moment \(\vec{p} \) in a uniform electric \(\overrightarrow{\mathbf{E}} \)?
(ii) What will happen if the field were not uniform? (Delhi 2017)
Answer:
(i)
(a) Torque on electric dipole. Consider an electric dipole consisting of two equal and opposite point charges separated by a small distance 2a having dipole moment

So net force on the dipole is zero
Since is uniform, hence the dipole does not undergo any translatory motion.

These forces being equal, unlike and parallel, from a couple, which rotates the dipole in clock-wise direction
∴ Magnitude of torque = Force × arm of couple

[The direction of \(\vec{\tau} \) is given by right hand screw rule and is normal to \(\vec{p} \)  ] and \(\vec{E} \)
Special cases
(i) when θ = 0 then τ = PE sin θ = 0
∴ Torque is zero and the dipole is in stable equilibrium
(ii) When θ = 90 then τ = PE sin 90 = PE
∴ The Torque is maximum
(b) Ratio of flux


∴ Electric flux through the sphere S₁ decreases with the introduction of dielectric inside it.

(ii) If the electric field is non uniform, the dipole experiences a translatory force as well as a torque.

Question 24.
State Gauss’s law in electrostatics. Derive an expression for the electric field due to an infinitely long straight uniformly charged wire. (Comptt. Delhi 2017)
Answer:
Gauss Theorem : The surface integral of electric field over a closed surface is equal to \(\frac { 1 }{ { \epsilon }_{ 0 } } \) times the charge enclosed by the surface.

Question 25.
Two small identical electrical dipoles AB and CD, each of dipole moment ‘p’ are kept at an angle of 120° as shown in the figure. What is the resultant dipole moment of this combination? If this system is subjected to the electric field (\(\vec{E}\)) directed along +X direction, what will be the magnitude and direction of the torque acting on this?

Answer:
The resultant dipole moment of the combi-nation is
P_R = \(\sqrt{p^{2}+p^{2}+2 p^{2} \cos 120^{\circ}}\) = p

since cos 120° = -1/2
This will make an angle of 30° with the X-axis, therefore torque acting on it is
τ=PE sin 30° = pE/2 (Along Z-direction)

Question 26.
A metallic spherical shell has an inner radius R₁ and outer radius R₂. A charge Q is placed at the center of the spherical cavity. What will be surface charge density on (i) the inner surface, and (ii) the outer surface?
Answer:
The induction of charges is as shown.

Therefore surface charge density on the inner and the outer shell is on the outer surface is

Question 27.
If the total charge enclosed by a surface is zero, does it imply that the electric field everywhere on the surface is zero? Conversely, if the electric field everywhere on a surface is zero, does it imply that the net charge inside is zero.
Answer:
No, the field may be normal to the surface. However, the converse is true i.e. when the electric field everywhere on the surface be zero then the net charge inside it must be zero.

Long Answer Type

Question 1.
(a) State Gauss theorem in electrostatics. Using it, prove that the electric field at a point due to a uniformly charged infinite plane sheet is independent of the distance from it.
Answer:
It states, “The net electric flux through any Gaussian surface is equal to \(\frac{1}{\varepsilon_{0}}\) times the net electric charge enclosed by the surface.

Mathematically, Φ =\( \phi \vec{E} \cdot d \vec{A}=\frac{q_{i n}}{\varepsilon_{0}}\)

Consider an infinite plane sheet of charge. Let a be the uniform surface charge density, i.e. the charge per unit surface area. From symmetry, we find that the electric field must be perpendicular to the plane of the sheet and that the direction of E on one side of the plane must be opposite to its direction on the other side as shown in the figure below. In such a case let us choose a Gaussian surface in the form of a cylinder with its axis perpendicular to the sheet of charge, with ends of area A.

The charged sheet passes through the middle of the cylinder’s length so that the cylinder’s ends are equidistant from the sheet. The electric field has a normal component at each end of the cylinder and no normal component along the curved surface of the cylinder. As a result, the electric flux is linked with only the ends and not the curved surface.

Therefore by the definition of eLectric flux, the flux Linked with the Gaussian surface is given by
Φ = \(\oint_{A} \vec{E} \cdot \vec{\Delta} A\)
Φ = E_A + E_A = 2E_A … (1)

But by Gauss’s Law
Φ = \(\frac{q}{\varepsilon_{0}}=\frac{\sigma A}{\varepsilon_{0}}\) [∵ q = σA] … (2)

From equations (1) and (2), we have
2E_A = \(\frac{\sigma A}{\varepsilon_{0}}\) … (3)
E = \(\frac{\sigma}{2 \varepsilon_{0}}\) …. (4)

This gives the electric field due to an infinite plane sheet of charge which is independent of the distance from the sheet.

(b) How is the field directed if (i) the sheet is positively charged, (ii) negatively charged?
Answer:
(a) directed outwards
(b) directed inwards.

Question 2.
Use Gauss’s law to derive the expression for the electric field (\(\vec{E}\) ) due to a straight uniformly charged infinite line of charge λ Cm^-1.
Answer:

Consider an infinitely Long, thin wire charged positively and having uniform Linear charge density λ. The symmetry of the charge distribution shows that must be perpendicular to the tine charge and directed outwards. As a result of this symmetry, we consider a Gaussian surface in the form of a cylinder with arbitrary radius r and arbitrary Length L. with its ends perpendicular to the wire as shown in the figure. Applying Gauss’s theorem to curved surface ΔA₁ and circular surface ΔA₂.

Φ EΔA1 cos 0°+ EΔA2 cos 90° = \(\frac{q}{\varepsilon_{0}}\) = \(\frac{\lambda l}{\varepsilon_{0}}\) [∵ λ = \(\frac{q}{e}\)]
Or
E . 2πrl = \(\frac{\lambda l}{\varepsilon_{0}}\) ⇒ E = \(\frac{1}{2 \pi \varepsilon_{0}} \frac{\lambda}{r}\)

This is the expression for the electric field due to an infinitely long thin wire.
The graph is as shown.

Question 3.
Obtain the expression for the potential energy of an electric dipole placed with its axis at an angle (θ) to an external electric field (E). What is the minimum value of the potential energy?
Answer:
The torque x acting on an electric dipole of dipole moment p placed in a uniform electric field E is given by:
τ = pEsinθ …(i)
where θ is the angle made by the dipole with the electric field E. The torque tends to align the dipole along the direction of the field. If the dipole is rotated through a small angle dθ against the torque, work has to be done, which is stored in the form of the potential energy of the dipole.

Work done in rotating the dipole through the angle dθ against the torque τ is given by dw = τ dθ = pE sin θ dθ

If the dipole is rotated from θ₁ to θ₂, then
Total work is done,

W = – pE(cos θ₂ – cos θ₁)

This work done is stored as potential energy U.

U = – pE(cos θ₂ – cos θ₁)
If the dipole is rotated from θ₁ = \(\frac{\pi}{2}\) to θ₂ =θ, then

U = – pE(cos θ – cos \(\frac{\pi}{2}\) )
= – pE(cos θ – 0)

U = – pE cos θ
minimum value of potential energy
U = – pE When θ = 0°

Question 4.
Why does the electric field inside a dielectric decrease when it is placed in an external electric field?
Answer:
When a dielectric is placed in an electric field (E₀), it gets polarised, i.e. within the dielectric, an electric field (E) is induced in a direction opposite to that of the external field. Therefore, the net field within the dielectric decreases to \(\vec{E}_{0}\) – \(\vec{E}\)

Question 5.
Two-point charges +q and -2q are placed at the vertices ‘B’ and ‘C’ of an equilateral triangle ABC of side ‘a’ as given in the figure. Obtain the expression for (i) the magnitude and (ii) the direction of the resultant electric field at vertex A due to these two charges.

Answer:
The electric fields due to the two charges placed at B and C are inclined at an angle θ = 120° as shown
Now in magnitude, we have

E_B = k\(\frac{q}{a^{2}}\) and
E_C = k\(\frac{2q}{a^{2}}\) = 2 E_B

Hence E = \(\sqrt{E_{B}^{2}+E_{c}^{2}+2 E_{B} E_{c} \cos \theta}\)
Or E = \(\sqrt{E_{B}^{2}+\left(2 E_{B}\right)^{2}+2 E_{B}\left(2 E_{B}\right) \cos 120^{\circ}}\)

On Solving we have
E = \(\sqrt{3} E_{B}\) = \(\sqrt{3} \frac{k q}{a^{2}}\)

Direction

Therefore B = 90°, the resultant is inclined at an angle of 90° with EB.

Question 6.
Four-point charges Q, q, Q, and q are placed at the corners of a square of side ‘a’ as shown in the figure.
Find the
(a) the resultant electric force on a charge Q and
(b) the potential energy of this system.
Answer:
(a) Let us find the force on charge Q at point C.

Force due to charge Q placed at point A is
F_A = k\(\frac{Q^{2}}{(a \sqrt{2})^{2}}\) =k\(\frac{Q^{2}}{2 a^{2}}\) along AC

Force due to the charge q placed at D
F_D = k\(\frac{q Q}{a^{2}}\) alongDA

Force due to the charge q placed at B
F_B = k \(\frac{q Q}{a^{2}}\) along BC

The resuLtant of F_D and F_B is
F_BD = K\(\frac{q Q \sqrt{2}}{a^{2}}\) along AC

∴ net force of charge Q placed at point C is

(b) PotentiaL energy of the system

Question 7.
A charge +Q is uniformly distributed within a sphere of radius R. Find the electric field, distant r from the center of the sphere where: (1) r < R and (2) r > R.
Answer:
For a solid sphere p = \( \frac{q}{\frac{4}{3} \pi R^{3}}\) = \(\frac{q}{\text { volume }}\)

Case 1. 0 < r < R The point Lies within the sphere.

Using Gauss’s theorem
Let Q’ be the charge encLosed by Gaussian’s surface of radius r < R.

E(4πr²)=\(\frac{Q^{\prime}}{\varepsilon_{0}}=\frac{Q^{\prime}}{4 \pi \varepsilon_{0} r^{2}}\)

From (i) and (ii)
E = \(\frac{Q r^{3}}{4 \pi \varepsilon_{0} r^{2} R^{3}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q r}{R^{3}}\)

Case 2. For r > R
\(\oint \vec{E} \cdot \overrightarrow{d s}\) = \(\frac{q_{\text {enclosed }}}{\varepsilon_{0}}\)

E(4πr²) = \(\frac{Q}{\varepsilon_{0}}\)
E = \(\frac{Q}{4 \pi \varepsilon_{0} r^{2}}\)

Question 8.
(a) An electric dipole is kept first to the left and then to the right of a negatively charged infinite plane sheet having a uniform surface charge density. The arrows p₁ and p₂ show the directions of its electric dipole moment in the two cases. Identify for each case, whether the dipole is in stable or unstable equilibrium. Justify each answer.
Answer:
p₁: stable equilibrium, p₂: unstable equilibrium. The electric field, on either side, is directed towards the negatively charged sheet and its magnitude is independent of the distance of the field point from the sheet. For position P₁ dipole moment and electric field are parallel. For position p₂, they are antiparallel.

(b) Next, the dipole is kept in a similar way (as shown), near an infinitely long straight wire having uniform negative linear charge density. Will the dipole be in equilibrium at these two positions? Justify your answer.

Answer:
The dipole will not be in equilibrium in any of the two positions.
The electric field due to an infinite straight charged wire is non-uniform (E ∝ 1/r).
Hence there will be a net non-zero force on the dipole in each case.

Question 9.
Two large parallel plane sheets have uniform charge densities +σ and -σ. Determine the electric field (i) between the sheets, and (ii) outside the sheets.
Answer:
Let us consider two parallel planes charged conductors A and B carrying +ve and -ve charge density σ (charge per unit area). According to Gauss’ theorem, the electric intensity at P due to the charge on sheet A is

EA = \(\frac{\sigma}{2 \varepsilon_{0}}\) (From A and B)

The electric field at P due to charge on sheet B is
E= \(\frac{\sigma}{2 \varepsilon_{0}}\) (From A to B)

The electric field at P is
E = EA + EB
= \(\frac{\sigma}{\varepsilon_{0}}\)

Question 10.
Define electric flux and write its SI unit. The electric field components in the figure shown are : E_x = αx, E_y = 0, E_z = 0 where α = 100N/cm. Calculate the charge within the cube, assuming a = 0.1 m.
Or
An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude 2.0 × 10⁴ N/C (figure a)

(a) Calculate the time it takes to fall through this distance starting from rest.

(b) If the direction of the field is reversed (figure b) keeping its magnitude unchanged, calculate the time taken by a proton to fall through this distance starting from rest.
Answer:
Electric Flux is the dot product of the electric field and area vector.
Φ = \(\oint \vec{E} \cdot \overrightarrow{d s}\)

SI Unit: Nm²/C or Vm

For a given case
Φ = Φ₁ – Φ₂ = [Ex(atx= 2a) – Ex (atx = a)]a²
= [α(2a)-α(a)]a² = αa³
= 10⁴ × (0.1 )³ = 10 Nm²/C
But
Φ = \(\frac{q}{\varepsilon_{0}}\)

∴ q = ε₀Φ = 8.854 × 10^-12 × 10 C = 8.54 pC
Or
We have
F = qE

Acceleration, a = \(\frac{q E}{m}\)

Also
s = \(\frac { 1 }{ 2 }\) at² [u = 0]
∴ t = \(\sqrt{\frac{2 s}{a}}\)

(i) For the electron
a = \(\frac { eE}{ m }\)

∴ t = \(\sqrt{\frac{2 s m}{e E}}\)

∴ t = \(\sqrt{\frac{3 \times 10^{-2} \times 9.1 \times 10^{-31}}{1.6 \times 10^{-19} \times 2.0 \times 10^{4}}}\) = 2.92 ns

(ii) for proton
t = \(\sqrt{\frac{2 \times 1.5 \times 10^{-2} \times 1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 2 \times 10^{4}}}\)
= -.125 μs

Question 11.
What will be the total flux through the faces of the cube (figure) with the side of length ‘a’ if a charge q is placed at
(a) A: a corner of the cube.
(b) B: mid-point of an edge of the cube.
(c) C: center of the face of the cube.
(d) D: mid-point of B and C. {NCERT Exemplar)

Answer:
(a) The charge wilt is shared by eight cubes if it has to be enclosed. Therefore the flux through the cube will be one-eighth of the total flux. Φ = q/8ε₀
(b) The charge will be shared by four cubes if it has to be enclosed. Therefore the flux through the cube will be one-fourth of the total flux. Φ = q/ 4ε₀
(c) The charge will be shared by two cubes if it has to be enclosed. Therefore the flux through the cube will be one-half of the total flux. Φ = q/ 2ε₀
(d) The charge will be shared by two cubes if it has to be enclosed. Therefore the flux through the cube will be one-half of the total flux. Φ = q/ 2ε₀

Question 12.
Two charges q and -3q are placed fixed on the x-axis separated by distance ‘d’. Where should a third charge 2q be placed such that it will not experience any force? (NCERT Exemplar)
Answer:
The situation is as shown.

Let the charge 2q be placed at a distance ‘x’ from charge q. For the charge 2q to experience zero force we have
\(\frac{2 q^{2}}{4 \pi \varepsilon_{0} x^{2}}=\frac{6 q^{2}}{4 \pi \varepsilon_{0}(d+x)^{2}}\)
(d + x)² = 3x²

Solving for x we have
x = \(\frac{d}{2} \pm \frac{\sqrt{3} d}{2}\)
(-ve sign would be between q and -3q and hence is unacceptable.)

Therefore, we have
x = \(\frac{d}{2}+\frac{\sqrt{3} d}{2}=\frac{d}{2}(1+\sqrt{3})\) to the left of q.

Question 13.
(a) State Gauss’s law. Use it to deduce the expression for the electric field due to a uniformly charged thin spherical shell at points (i) inside and (ii) outside the shell. Plot a graph showing the variation of the electric field as a function of r > R and r < R. (r being the distance from the center of the shell)
Answer:
Gauss’s law states that the net outward flux through any closed surface is equal to 1 /ε₀ times the charge enclosed by the closed surface.

Consider a thin spherical shell of radius R and center at O. Let σ be the uniform surface charge density (charge per unit surface area) and q be the total charge on it. The charge distribution is spherically symmetric. Three cases arise

Case 1: at a point outside the spherical shell
In order to find the electric field at a point P outside the shell let us consider a Gaussian surface in the form of a sphere of radius r (r >>R).

By symmetry, we find that the electric field acts radially outwards and has a normal component at alt points on the Gaussian sphere. Therefore by definition of electric flux we have

Φ = E × A, where A is the surface area of the Gaussian sphere therefore
Φ = E × 4πr² …(1)

But by Gauss’s law
Φ = \(\frac{Q}{\varepsilon_{0}}=\frac{\sigma A}{\varepsilon_{0}}=\frac{\sigma \times 4 \pi R^{2}}{\varepsilon_{0}}\) … (2)

from equations (1) and (2) it follows that
E × 4πr² = \(\frac{Q}{\varepsilon_{0}}\) Or E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r^{2}}\)

E × 4πr² = \(\frac{\sigma \times 4 \pi R^{2}}{\varepsilon_{0}}\)

Ē = \(\frac{\sigma R^{2}}{\varepsilon_{0} r^{2}}\) … (3)

It follows that the electric field due to a spherical shell outside it is same as that due to a point charge. Therefore for points Lying outside the spherical shell the shell behaves as if the entire charge is concentrated at the centre.

Case 2: at a point inside the spherical shell
In this case, the Gaussian surface Lies inside the shell. Since no charge is enclosed In this surface therefore we have

E × 4πr²=q₀/ε₀ [∵ q=0]
Therefore E = 0

(b) Two identical metallic spheres A and B having charges +4Q. and -10 Q are kept a certain distance apart. A third identical uncharged sphere C is first placed in contact with sphere A and then with sphere B. Spheres A and B are then brought in contact and then separated. Find the charges on the spheres A and B.
Answer:
The initial charge on the sphere A = + 4 Q.
The initial charge on the sphere B = -10 Q.

Since all the three spheres are identical, i.e. they have the same capacity, therefore when uncharged sphere C is placed in contact with A, the total charge is equally shared between them.

Charge on C after contact with A
= \(\frac{0+4 Q}{2}\) = 2Q

Charge on A after contact with C is 2Q.
When sphere C carrying a charge 2Q is placed in contact with B, again charges are equally shared between C and B equally.

Charge on C after it is in contact with B
= \(\frac{2 Q-10 Q}{2}\) = -4Q

Now when sphere A with a charge of 2Q. is placed in contact with B, with charge -4Q.

Charge are again shared
∴ charge on A or B = \(\frac{2 Q-4 Q}{2}\) = -Q.

Question 14.
(a) Define electric dipole moment. Is it a scalar or a vector? Derive the expression for the electric field of a dipole at a point on the equatorial plane of the dipole.
Answer:
It is defined as the product of the magnitude of either of the two charges and the distance between them.
For derivation see sol. 9(a) of LA-II.

(b) Draw the equipotential surfaces due to an electric dipole. Locate the points where the potential due to the dipole is zero.
Answer:
The diagram is as shown

The zero potential points lie on the equatorial line.

Question 15.
(a) Using Gauss’s law obtain expressions for the electric field (i) inside, and (ii) outside a positively charged spherical shell.
(c) A Square plane sheet of side 10 cm is inclined at an angle of 30° with the direction of a uniform electric field of 200 NC“1. Calculate the electric flux passing through the sheet.
Answer:
(a) Spherical Shell
Consider a spherical shell of radius R. Let q be a charge on the shell. Let us find the electric field at a point P at a distance r from the center 0 of the spherical shell.

Case (i): When point P lies inside the spherical shell
From the point, P draws a Gaussian surface which will be a sphere of radius r.

From the Gauss’s Theorem,
\(\oint_{s} \vec{E} \cdot \vec{d} S=\frac{0}{\varepsilon_{0}}\) [∵ No charge exists inside the spherical shell]

Or
E = 0
i. e. electric field inside the charged spherical shell is zero.

Case (ii): When point P is lying outside the shell (i.e. r > R)
From point P, draw a Gaussian surface which will be a spherical shell of radius r. Let dS be a small area element on the Gaussian surface P.


Or

i. e. the electric field outside the spherical shell behaves as if the whole charge is concentrated at the center of the spherical shell.

(b) Show graphically variation of the electric field as a function of the distance r from the center of the sphere.
Answer:
Variation of electric field E with distance
The given figure shows the variation of an electric field with distance from the center of the charged spherical shell.

(c) Here E = 200 N/C, S = 0.1 × 0.1 = 0.01 m²
And θ = 90° – 30° = 60°

The electric flux linked with the square sheet
Φ = E S cos 60°
= 200 × 0.01 × \(\frac { 1 }{ 2 }\) = 1.0 Nm² C^-1

Question 16.
A charge is distributed uniformly over a ring of radius ‘a’. Obtain an expression for the electric intensity E at a point on the axis of the ring. Hence show that for points at a large distance from the ring, it behaves like a point charge.
Answer:
Consider a uniformly charged ring of radius ‘a’. Let the total charge on the ring be Q, Let us find the electric field on the axis of the ring at point P distance x from the center of the ring. Consider a segment of charge dQ as shown in the figure.

The magnitude of etectñc field at P due to the segment is
dE=k\(\frac{d Q}{r^{2}}\) …(1)

This field can be resolved into its components: x component dE_x = dE cos α an along the axis of the ring and y component dE perpendicular to the axis. Since these perpendicular components, due to alt the charge segments, are equal and opposite, therefore they cancel out each other. From the diagram we have r = \(\sqrt{x^{2}+a^{2}}\) and cos α = x/r, therefore we have
…(2)

In this case, all the segments of the ring give the same contribution to the field at P since they are all equidistant from this point. Thus we can easily sum over all segments to get the total electric field at point P

If the point of observation is far away, i.e. x >> a, then E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{x^{2}}\). This is the same as that for a point charge. Thus at far-off axial points, a charged ring behaves as if a point charge is situated at the center of the ring.

Question 17.
Two thin concentric and coplanar spherical shells, of radii ‘a’ and ‘b’ (b > a), carry charges, q, and Q respectively. Find the magnitude of the electric field, at a point distant x, from their common center for
(i) O < x < a
(ii) a ≤ x < b
(iii) b ≤ x < ∞
Answer:
The diagram is as shown.

(i) For 0 < x < a
Point Lies inside both the spherical shells.

∴ charge enclosed = 0
Hence, E(x) = 0

(ii) For a < x < b
Point is outside the spherical shell of radius ‘a’ but inside the spherical Shell of radius ‘b’.
Therefore

(iii) For b ≤ x< ∞
The point is outside of both the spherical shells. The total effect we charge at the center equals (Q + q).

Question 18.
(a) Define electric flux. Is It a scalar or a vector quantity?

A point charge q is at a distance of d/2 directly above the center of a square of side d, as shown In the figure. Us. Gauss’s law to obtain the expression for the electric flux through the square.
Answer:
The electric flux through a given surface is defined as the dot product of the electric field and area vector over that surface. It is a scalar quantity.

Constructing a cube of side ‘d’ so that charge ‘q’ gets placed within this cube (Gaussian surface)
According to Gauss’s law the electric flux
Φ = \(\frac{q}{\varepsilon_{0}}\)

This is the total flux through all the six faces of the cube.

Hence electric flux through the square (one face of the cube)
Φ = \(\frac{1}{6} \frac{q}{\varepsilon_{0}}\)

(b) If the point charge is now moved to a distance ‘d’ from the center of the square and the side of the square is doubled, explain how the electric flux will be affected.
Answer:
If the charge is moved to a distance ‘d’ and the side of the square is doubled the cube will be constructed to have a side 2d but the total charge enclosed in it will remain the same. Hence the total flux through the cube and therefore the flux through the square will remain the same as before.

Question 19.
(a) Draw a graph to show the variation of E with perpendicular distance r from the line of charge.
Answer:
The required graph is

(b) Find the work done in bringing a charge q from perpendicular distance r₁ to r₂ (r₂ > r₁).
Answer:
Work done in moving the charge “q”. through a smaLL displacement ‘dr’

dW = \(\vec{F}\) . \(\vec{dr}\) =q\(\vec{E}\) .\(\vec{dr}\)
dW = qE dr cos 0 = qEdr
dW= q x \(\frac{\lambda}{2 \pi \varepsilon_{0} r}\) dr

Hence work done in moving the charge from r₁ to r₂ (r₂ > r₁)

Question 20.
Derive an expression for the torque acting on an electric dipole of dipole moment \(\vec{p}\) placed in a uniform electric field \(\vec{E}\). Write the direction along which the torque acts.
OR
Derive an expression for the electric field at a point on the axis of an electric dipole of dipole moment \(\vec{p}\). Also, write its expression when the distance r >> a the length ‘a’ of the dipole.
Answer:
Consider an electric dipole consisting of charges -q and +q and dipole length d is placed in a uniform electric field E as shown in the figure.

Let the dipole moment makes an angle θ with the direction of the electric field. The two charges experience force qE each.

These forces are equal, parallel, and opposite. Therefore, the net force acting on the dipole is
F_n = qE – qE = 0

Thus the net force acting on the dipole is zero.

These two forces constitute a couple. This applies a torque on the dipole given by
τ = either force × arm of the couple
τ = qE × d sin θ, where d sin θ is the arm of the couple.
τ = q d Esin θ, where p = qd, dipole moment
τ = pE sin θ.

Consider an electric dipole consisting of -q and +q charges separated by a distance 2a as shown in the figure. Let P be the point of observation on the axial Line where the electric field has to be found. Let it be at a distance r from the center O of the dipole. Let us suppose that the dipole is placed in a vacuum.

Let E and E_B be the electric fields at point P due to the charges at A and B respectively. Therefore,

The two fields at P are in opposite directions. Thus the resultant electric field at P is given by
E = \(\sqrt{E_{A}^{2}+E_{B}^{2}-2 E_{A} E_{B} \cos \theta}=\sqrt{\left(E_{B}-E_{A}\right)^{2}}\) = E_B – E_A

since θ = 180⁰ Therefore, the resultant electric field is

Solving we have

where p = q 2a.
If the dipole is short then r >> a, therefore, ‘a’ is neglected as compared to r, hence

This field is in the direction of the dipole moment.

Question 21.
(a) Derive the expression for the electric field at a point on the equatorial line of an electric dipole.
Answer:
Consider an electric dipole consisting of charges -q and +q separated by a distance 2a as shown in the figure. Let the point of observation P lie on the. right bisector of the dipole AB at a distance r from its midpoint 0. Let E_A and E_B be the electric field intensities at point P due to charges at A and B, respectively.

The two electric fields have magnitudes

in the direction of AP

in the direction of PB.

The two fields are equal in magnitude but have different directions. Resolving the two fields E_A and E_B into their rectangular components, i.e. perpendicular to and parallel to AB. The components perpendicular to AB, i.e E_A sinG and E_B sinG being equal and opposite cancel out each other while the components parallel to AB, i.e. E_A cos θ and E_B cos θ being in the same direction add up as shown in the figure. Hence the resultant electric field at point P is given by

E = E_Acosθ + E_Bcosθ

∵ cos θ = \(\frac{a}{\left(r^{2}+a^{2}\right)^{1 / 2}}\) and q 2a = p

For a short dipole r²,>>a² therefore
E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{p}{r^{3}}\)

(b) Discuss the orientation of the dipole in (a) stable, (b) unstable equilibrium in a uniform electric field.
Answer:
For stable equilibrium \(\vec{P}\) is along \(\vec{E}\)
θ = 0°, τ = PE sin 0°, Torque alligns the dipole in the direction of field

For unstable equilibrium \(\vec{P}\) is antiparallel to \(\vec{E}\)
∵ θ = 180°, τ = PE sin 180° = 0, Torque alligns the dipole in a direction opposite to \(\vec{E}\).

Question 22.
(a) An electric dipole of dipole moment \(\vec{p}\) consists of point charges +q and -q separated by a distance 2a apart. Deduce the expression for the electric field due to the dipole at a distance x from the center of the dipole on its axial line in terms of the dipole moment. Hence show that in

the limit r >> a, \(\vec{E}\) → \(\frac{2 \vec{p}}{4 \pi \varepsilon_{0} r^{3}}\)
Answer:

Consider an electric dipole consisting of charges -q and +q and dipole length d is placed in a uniform electric field E as shown in the figure.

Let the dipole moment makes an angle θ with the direction of the electric field. The two charges experience force qE each.

These forces are equal, parallel, and opposite. Therefore, the net force acting on the dipole is
F_n = qE – qE = 0

Thus the net force acting on the dipole is zero.

These two forces constitute a couple. This applies a torque on the dipole given by
τ = either force × arm of the couple
τ = qE × d sin θ, where d sin θ is the arm of the couple.
τ = q d Esin θ, where p = qd, dipole moment
τ = pE sin θ.

Consider an electric dipole consisting of -q and +q charges separated by a distance 2a as shown in the figure. Let P be the point of observation on the axial Line where the electric field has to be found. Let it be at a distance r from the center O of the dipole. Let us suppose that the dipole is placed in a vacuum.

Let E and EB be the electric fields at point P due to the charges at A and B respectively. Therefore,

The two fields at P are in opposite directions. Thus the resultant electric field at P is given by
E = \(\sqrt{E_{A}^{2}+E_{B}^{2}-2 E_{A} E_{B} \cos \theta}=\sqrt{\left(E_{B}-E_{A}\right)^{2}}\) = E_B – E_A

since θ = 1800 Therefore, the resultant electric field is

Solving we have

where p = q 2a.
If the dipole is short then r >> a, therefore, ‘a’ is neglected as compared to r, hence

This field is in the direction of the dipole moment.

(b) Given the electric field in the region E = 2xî, find the net electric flux through the cube and the charge enclosed by it.

Answer:
Since the electric field has only an x component, for faces perpendicular to the x-direction, the angle between E and ΔS is ± π/2.

Therefore, the flux Φ = E.ΔS is separately zero for each face of the cube except the two faces along the X-axis.

Now the magnitude of the electric field at the left face is
E_L= 0 (x = 0 at the left face)

The magnitude of the electric field at the right face is
E_R = 2x = 2a (x = a at the right face)

The corresponding fluxes are
Φ_L = E_L.ΔS = ΔS (\(\vec{E}_{L} \cdot \hat{n}_{\mathrm{L}}\)) = E_LΔS COS θ
= – E_L ΔS = 0, since θ =180°

Φ_R = E_R.ΔS = E_RΔS cos θ = E ΔS = (2a)a², since θ = 0°

Net flux through the cube
Φ = -Φ_R + Φ_L = 2a³ -0 = 2a³

We can use Gauss’s law to find the total charge q inside the cube.
We have Φ = q/ε₀ or q = Φε₀. Therefore, q = 2a³ × 8.854 × 10^-12C

Question 23.
(a) Derive an expression for the electric field at any point on the equatorial line of an electric dipole.
Answer:
Consider an electric dipole consisting of charges -q and +q separated by a distance 2a as shown in the figure. Let the point of observation P lie on the. right bisector of the dipole AB at a distance r from its midpoint 0. Let EA and EB be the electric field intensities at point P due to charges at A and B, respectively.

The two electric fields have magnitudes

in the direction of AP

in the direction of PB.

The two fields are equal in magnitude but have different directions. Resolving the two fields E_A and E_B into their rectangular components, i.e. perpendicular to and parallel to AB. The components perpendicular to AB, i.e E_A sinG and E_B sinG being equal and opposite cancel out each other while the components parallel to AB, i.e. E_A COS θ and E_B cos θ being in the same direction add up as shown in the figure. Hence the resultant electric field at point P is given by

E = E_A cos θ + E_B cos θ

∵ cos θ = \(\frac{a}{\left(r^{2}+a^{2}\right)^{1 / 2}}\) and q 2a = p

For a short dipole r²,>>a² therefore
E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{p}{r^{3}}\)

(b) Two identical point charges, q each, are kept 2 m apart in the air. A third point charge Q of unknown magnitude and sign is placed on the line joining the charges such that the system remains in equilibrium. Find the position and nature of Q.
Answer:
The third charge Q. wilt is in equilibrium if it experiences zero net force. Let it be placed at a distance x meter from the charge q.

Solving for x, we have x= 1 m

For the equilibrium of charges “q”, the nature of charge Q must be opposite to the nature of charge q and should be placed at the center of two charges.

Question 24.
(a) Define electric flux. Write its S.I. unit.
Answer:
It is defined as the total number of electric field lines crossing a given area. The electric flux can be found by multiplying the component of the electric field in the direction of the area vector (or perpendicular to the area) with the area of the closed surface. It is measured in Nm²C-^-1.

(b) A small metal sphere carrying charge +Q is located at the center of a spherical cavity inside a large uncharged metallic spherical shell as shown in the figure. Use Gauss’s law to find the expressions for the electric field at points P₁ and P₂.

Answer:
For point P, using Gauss law we have

Since E and dA are in the same direction therefore we have E = \(\frac{Q}{\varepsilon_{0} A}\)

Point P₂ lies inside the metal, therefore the Gaussian surface drawn at P₂ does not include a charge, hence the electric field at P₂ is zero.

(c) Draw the pattern of electric field lines in this arrangement.
Answer:
The electric field lines are as shown.

Question 25.
(a) Deduce the expression for the torque acting on a dipole of dipole moment \(\vec{P}\) in the presence of a uniform electric field \(\vec{E}\)
Answer:

Consider an electric dipole consisting of charges -q and +q and dipole length d is placed in a uniform electric field E as shown in the figure.

Let the dipole moment makes an angle θ with the direction of the electric field. The two charges experience force qE each.

These forces are equal, parallel, and opposite. Therefore, the net force acting on the dipole is
F_n = qE – qE = 0

Thus the net force acting on the dipole is zero.

These two forces constitute a couple. This applies a torque on the dipole given by
τ = either force × arm of the couple
τ = qE × d sin θ, where d sin θ is the arm of the couple.
τ = q d Esin θ, where p = qd, dipole moment
τ = pE sin θ.

Consider an electric dipole consisting of -q and +q charges separated by a distance 2a as shown in the figure. Let P be the point of observation on the axial Line where the electric field has to be found. Let it be at a distance r from the center O of the dipole. Let us suppose that the dipole is placed in a vacuum.

Let E and EB be the electric fields at point P due to the charges at A and B respectively. Therefore,

The two fields at P are in opposite directions. Thus the resultant electric field at P is given by
E = \(\sqrt{E_{A}^{2}+E_{B}^{2}-2 E_{A} E_{B} \cos \theta}=\sqrt{\left(E_{B}-E_{A}\right)^{2}}\) = E_B – E_A

since θ = 180° Therefore, the resultant electric field is

Solving we have

where p = q 2a.
If the dipole is short then r >> a, therefore, ‘a’ is neglected as compared to r, hence

This field is in the direction of the dipole moment.

(b) Consider two hollow concentric spheres S₁ and S₂ and enclosing charges 2Q and 4Q. respectively as shown in the figure, (i) Find out the ratio of the electric flux through them, (ii) How will the electric flux through the sphere S₁ change if a medium of dielectric constant ‘er’ is introduced in the space inside S1 in place of air? Deduce the necessary expression.

Answer:
Φ₁ = \(\frac{2 Q}{\varepsilon_{0}}\) and Φ2 = \(\frac{2 Q+4 Q}{\varepsilon_{0}}=\frac{6 Q}{\varepsilon_{0}}\)

Hence ratio

When a dielectric of dielectric constant εr is introduced then we have
Φ₁ = \(\frac{2 Q}{\varepsilon}=\frac{2 Q}{\varepsilon_{r} \varepsilon_{0}}\)

Question 26.
(a) Using Gauss’ law, derive an expression for the electric field intensity at any point outside a uniformly charged thin spherical shell of radius R and charge density a C/m2. Draw the field lines when the charge density of the sphere is
(i) positive,
(ii) negative.
(b) A uniformly charged conducting sphere of 2.5 m in diameter has a surface charge density of 100 µC/m². Calculate the
(i) charge on the sphere
(ii) total electric flux passing through the sphere (Delhi 2008)
Answer:
(a) (i) To find out electric field at a point outside a spherical charged shell we imagine a symmetrical Gaussian surface in such a way that the point lies on it.

Question 27.
(a) Define electric flux. Write its SI units.
(b) The electric field components due to a charge inside the cube of side 0.1 m are as shown :
E_x = ax, where α = 500 N/C^-m

Calculate
(i) the flux through the cube, and
(ii) the charge inside the cube. (All India 2008)
Answer:
(a) Electric flux through a surface represents the total number of electric lines of force crossing the surface.
∴ S.I. unit is Nm² C^-1.
(b) (i) Flux through R.H.S. of the cube is

Question 28.
(a) Define electric flux. Write its S.I. units.
(b) Using Gauss’s law, prove that the electric field at a point due to a uniformly charged infinite plane sheet is independent of the distance from it.
(c) How is the field directed if
(i) the sheet is positively charged,
(ii) negatively charged? (Delhi 2012)
Answer:
(a) Electric flux: The electric flux through a given area held inside an electric field is the measure of the total number of electric lines of force passing normally through that area.

(b) Consider a thin, infinite plane sheet of charge with uniform surface charge density 0. We wish to calculate its electric field at a point P at distance r from it.

By symmetry, electric field E points outwards normal to the sheet. Also, it must have same magnitude and opposite direction at two points P anel F equidistant from the sheet and on opposite sides. We choose cylindrical Gaussian surface of cross¬sectional area A and length 2r with its axis perpendicular to the sheet.

As the lines of force are parallel to the curved surface of the cylinder, the flux through the curved surface is zero. The flux through the plane-end faces of the cylinder is :

(i) For positively charged sheet ➝ away from the sheet
(ii)For negatively charged sheet ➝towards the sheet

Question 29.
(a) Define electric flux. Write its S.I. unit.
(b) A small metal sphere carrying charge +Q is located at the centre of a spherical cavity inside a large uncharged metallic spherical shell as shown in the figure the expressions for the electric field at points P₁ and P₂.
(c) Draw the pattern of electric field lines in this arrangement. (Comptt. All India 2012)

Answer:
(a) Electric flux: The electric flux through a given area held inside an electric field is the measure of the total number of electric lines of force passing normally through that area.

(b) Consider a thin, infinite plane sheet of charge with uniform surface charge density 0. We wish to calculate its electric field at a point P at distance r from it.

By symmetry, electric field E points outwards normal to the sheet. Also, it must have same magnitude and opposite direction at two points P anel F equidistant from the sheet and on opposite sides. We choose cylindrical Gaussian surface of cross-sectional area A and length 2r with its axis perpendicular to the sheet.

As the lines of force are parallel to the curved surface of the cylinder, the flux through the curved surface is zero. The flux through the plane-end faces of the cylinder is :

(i) For positively charged sheet ➝ away from the sheet
(ii)For negatively charged sheet ➝towards the sheet

(b) Calculation of electric field at point Pt:
Net charge enclosed by the Gaussian surface is +Q

As electric field of positive charge is radially outwards, it is parallel to the area vector on the surface chosen.

As point P₂ lies inside the metal, therefore electric field at point P₂ is zero.

Question 30.
Define electric dipole moment. Is it a scalar or a vector? Derive the expression for the electric field of a dipole at a point on the equatorial
plane of the dipole. (All India 2013)
Answer:
Electric dipole moment: It is the product of the magnitude of either charge and distance between them.

It is a vector quantity whose direction is from negative to positive charge.

Electric field intensity at P due to +q charge is

Electric field intensity at P due to -q charge is,

Question 31.
Using Gauss’ law deduce the expression for the electric field due to a uniformly charged spherical conducting shell of radius R at a point
(i) outside and
(ii) inside the shell.
Plot a graph showing variation of electric field as a function of r > R and r < R (r being the distance from the centre of the shell) (All India)
Answer:
(i) Field Outside Shell :
Consider a thin spherical shell of radius R with centre O. Let charge +q be distributed uniformly over the surface of shell. To calculate electric field intensity at P where OP = r, imagine a sphere S, with centre at O and radius r. The surface of sphere is Gaussian surface over at every point. Electric field is same and directed radially outwards.

Applying Gauss’ theorem

\(\overrightarrow { r } \) is distance of point P from centre where E is calculated]
(ii) Inside Shell: As we know charge is located on its surface,

(iii) at r < R \(\overrightarrow { E } \) is zero and r  = R, E is maximum at r > R, E is decreasing at E ∝ \(\frac { 1 }{ { r }^{ 2 } } \)

Question 32.
Using Gauss’s law, derive the expression for the electric field at a point
(i) outside and
(ii) inside a uniformly charged thin spherical shell. Draw a graph showing electric field E as a function of distance from the centre.
Answer:
Electric field due to a uniformly charged spherical shell:
Suppose a thin spherical shell of radius R and centre O
Let the charge + q be distributed over the surface of sphere
Electric field intensity \(\overrightarrow { E } \) is same at every point on the surface of sphere directed directly outwards
Let a point P be outside the shell with radius vector

Question .
(a) Deduce the expression for the torque acting on a dipole of dipole moment \(\overrightarrow { p } \) in the presence of a uniform electric field E.
(b) Consider two hollow concentric spheres, S₁ and S₂, enclosing charges 2Q and 4Q respectively as shown in the figure.

(i) Find out the ratio of the electric flux through them.
(ii) How will the electric flux through the sphere S₁ change if a medium of dielectric constant ‘ε_r‘ is introduced in the space inside S₁, in place of air? Deduce the necessary expression. (All India 2014)
Answer:
(a) Torque on electric dipole. Consider an electric dipole consisting of two equal and opposite point charges separated by a small distance 2a having dipole moment

So net force on the dipole is zero
Since is uniform, hence the dipole does not undergo any translatory motion.

These forces being equal, unlike and parallel, from a couple, which rotates the dipole in clock-wise direction
∴ Magnitude of torque = Force × arm of couple

[The direction of \(\vec{\tau} \) is given by right hand screw rule and is normal to \(\vec{p} \)  ] and \(\vec{E} \)
Special cases
(i) when θ = 0 then τ = PE sin θ = 0
∴ Torque is zero and the dipole is in stable equilibrium
(ii) When θ = 90 then τ = PE sin 90 = PE
∴ The Torque is maximum
(b) Ratio of flux


∴ Electric flux through the sphere S₁ decreases with the introduction of dielectric inside it.

Question 33.
(a) An electric dipole of dipole moment \(\overrightarrow { p } \) consists of point charges + q and – q separated by a distance 2a apart. Deduce the expression for the electric field \(\overrightarrow { E } \) due to the dipole at a distance x from the centre of the dipole on its axial line in terms of the dipole moment \(\overrightarrow { p } \)?. Hence show that in the limit x >> a, \(\overrightarrow { E } \) —>2\(\overrightarrow { p } \) (4πε₀x³).
(Delhi 2015)
Answer:
(a) Expression for magnetic field due to dipole on its axial lane:

(b)Only the faces perpendicular to the direction of x-axis, contribute to the Electric flux. The remaining faces of the cube given zero

Question 34.
(a) Define electric flux. Write its S.I. unit. “Gauss’s law in electrostatics is true for any closed surface, no matter what its shape or size is”. Justify this statement with the help of a suitable example.
(b) Use Gauss’s law to prove that the electric field inside a uniformly charged spherical shell is zero. (All India)
(a) Electric flux. The electric lines of force passing through that area, when held normally to the lines of force.
Answer:
Electric flux. The electric lines of force passing through that area, when held normally to the lines of force.

S.I. units: Vm, Nm²C^-1
Gauss’s Law states that the electric flux through a closed surface is given by

The law implies that the total electric flux through a closed surface depends on the quantity of total charge enclosed by the surface, and does not depend on its shape and size.

For example, net charge enclosed by the electric dipole (q, -q) is zero, hence the total elec¬tric flux enclosed by a surface containing electric dipole is zero.
(b) Electrical field inside a uniformly charged spherical shell. Let us consider a point ‘P’ inside the shell. The Gaussian surface is a sphere through P centred at O.

The flux through the Gaussian surface is E × 4πr².

However, in this case, the Gaussian surface encloses no charge. Gauss’s law then gives
E × 4πr² = 0
or E = 0
(r < R)
that is, the field due to a uniformly charged thin shell is zero at all points inside the shell.

Question 35.
(a) Derive the expression for the energy stored in a parallel plate capacitor. Hence obtain the expression for the energy density of the electric field.
(b) A fully charged parallel plate capacitor is connected across an uncharged identical capacitor. Show that the energy stored in the combination is less than that stored initially in the single capacitor. (All India 2015)
Answer:
(a) (i) Energy of a parallel plate capacitor. Consider a capacitor of capacitance C. Initial charge on plates is zero. Initial potential difference between the capacitor plates is zero. Let a charge Q be given to it in small steps. When charge is given to capacitor, the potential difference between its plates increases. Let at any instant when charge on capacitor be q, the potential difference between its plates be,

Now work done in giving an additional infinitesimal charge dq to capacitor

The total work done in giving charge from 0 to Q will be equal to the sum of all such infinitesimal works, which may be obtained by integration. Therefore total work


If V is the final potential difference between capacitor plates, then Q = CV

This work is stored as electrostatic potential energy of capacitor i.e., Electriostatic potential energy,

(ii) Expression for Energy Density.
Consider a parallel plate capacitor consisting of plates, each of area A, separated by a distance d. If space between the plates is filled with a medium of dielectric constant K, then
Capacitence of Capacitor,

If σ is the surface charge density of plates, then electric field strength between the plates.

This is the expression for electrostatic energy density in medium of dielectric constan K. In air of free space  (K = 1), therefore energy density,

(b) The energy of the capacitor when fully charged is

When this charged capacitor is connected to an identical capacitor C, then the charge will be distributed equally, \(\frac { q }{ 2 } \) on each of the capacitors, then


Hence, the total energy stored is half of that stored initially in one capacitor which means the energy stored in combination is less than that stored initially in the single capacitor.

Question 36.
(i) Use Gauss’s law to find the electric field due to a uniformly charged infinite plane sheet. What is the direction of field for positive and negative charge densities?
(ii) Find the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors Cj and C2 with their capacitances in the ratio 1 : 2 so that the energy stored in the two cases becomes the same. (All India 2016)
Answer:
(i)
(a) Electric flux: The electric flux through a given area held inside an electric field is the measure of the total number of electric lines of force passing normally through that area.

(b) Consider a thin, infinite plane sheet of charge with uniform surface charge density 0. We wish to calculate its electric field at a point P at distance r from it.

By symmetry, electric field E points outwards normal to the sheet. Also, it must have same magnitude and opposite direction at two points P anel F equidistant from the sheet and on opposite sides. We choose cylindrical Gaussian surface of cross¬sectional area A and length 2r with its axis perpendicular to the sheet.

As the lines of force are parallel to the curved surface of the cylinder, the flux through the curved surface is zero. The flux through the plane-end faces of the cylinder is :

(i) For positively charged sheet ➝ away from the sheet
(ii)For negatively charged sheet ➝towards the sheet

(b) and

Question 37.
(a) Derive an expression for the electric field E due to a dipole of length ‘2a’ at a point distant r from the centre of the dipole on the axial line. (b) Draw a graph of E versus r for r >> a.
(c) If this dipole were kept in a uniform external electric field diagrammatically represent the position of the dipole in stable and unstable equilibrium and write the expressions for the torque acting on the dipole in both the cases. (Outside Delhi 2017)
Answer:
(a) Expression for magnetic field due to dipole on its axial lane:

(b)Only the faces perpendicular to the direction of x-axis, contribute to the Electric flux. The remaining faces of the cube given zero

(b) Graph between E Vs r

(i) Diagrammatic representation
(ii) Torque acting on these cases
(i) In stable equilibrium, torque is zero (θ = 0)

(ii) In unstable equilibrium also, torque is zero (θ = 180°)

Question 38.
(a) Use Gauss’s theorem to find the electronic field due to a uniformly charged infinitely large plane thin sheet with surface charge density a.
(b) An infinitely large thin plane sheet has a uniform surface charge density +a. Obtain the expression for the amount of work done in bringing a point charge q from infinity to a point, distant r, in front of the charged plane sheet. (Outside Delhi 2017)
Answer:
(a) Electric flux: The electric flux through a given area held inside an electric field is the measure of the total number of electric lines of force passing normally through that area.

(b) Consider a thin, infinite plane sheet of charge with uniform surface charge density 0. We wish to calculate its electric field at a point P at distance r from it.

By symmetry, electric field E points outwards normal to the sheet. Also, it must have same magnitude and opposite direction at two points P anel F equidistant from the sheet and on opposite sides. We choose cylindrical Gaussian surface of cross¬sectional area A and length 2r with its axis perpendicular to the sheet.

As the lines of force are parallel to the curved surface of the cylinder, the flux through the curved surface is zero. The flux through the plane-end faces of the cylinder is :

(i) For positively charged sheet ➝ away from the sheet
(ii)For negatively charged sheet ➝towards the sheet

Question 39.
(a) State Gauss’ law. Using this law, obtain the expression for the electric field due to an infinitely long straight conductor of linear charge density X.
(b) A wire AB of length L has linear charge density λ = kx, where x is measured from the end A of the wire. This wire is enclosed by a Gaussian hollow surface. Find the expression for the electric flux through this surface.
(Comptt. Outside Delhi 2017)
Answer:
(a)
Gauss’s law in electrostatics : It states that “the total electric flux over the surface S in vaccum is \(\frac{1}{\varepsilon_{0}}\) times the total charge (q).”

Electric field due to an infinitely long straight wire : Consider an infinitely long straight line charge having linear charge density X to determine its electric field at distance r. Consider a cylindrical Gaussian surface of radius r and length l coaxial with the charge. By symmetry, the electric field E has same magnitude at each point of the curved surface Sj and is directed radially outward.
Total flux through the cylindrical surface,

(b) Given : Length of wire = L, Charge density (λ) = kx, ϕ = ?
We know

Question 40.
Two point charges 4 (J.C and +1 pC are separated by a distance of 2 m in air. Find the point on the line-joining charges at which the net electric field of the system is zero. (Comptt. Outside Delhi 2017)
Answer:
q₁ = 4 µC, q₂ = 1 µC, r = 2 m
At this point, the net electric field of the system is zero.

Question 41.
(a) Use Gauss’s theorem to find the electric field due to a uniformly charged infinitely large plane thin sheet with surface charge density σ.
Answer:
It states, “The net electric flux through any Gaussian surface is equal to \(\frac{1}{\varepsilon_{0}}\) times the net electric charge enclosed by the surface.

Mathematically, Φ =\( \phi \vec{E} \cdot d \vec{A}=\frac{q_{i n}}{\varepsilon_{0}}\)

Consider an infinite plane sheet of charge. Let a be the uniform surface charge density, i.e. the charge per unit surface area. From symmetry, we find that the electric field must be perpendicular to the plane of the sheet and that the direction of E on one side of the plane must be opposite to its direction on the other side as shown in the figure below. In such a case let us choose a Gaussian surface in the form of a cylinder with its axis perpendicular to the sheet of charge, with ends of area A.

The charged sheet passes through the middle of the cylinder’s length so that the cylinder’s ends are equidistant from the sheet. The electric field has a normal component at each end of the cylinder and no normal component along the curved surface of the cylinder. As a result, the electric flux is linked with only the ends and not the curved surface.

Therefore by the definition of eLectric flux, the flux Linked with the Gaussian surface is given by
Φ = \(\oint_{A} \vec{E} \cdot \vec{\Delta} A\)

Φ = E_A + E_A = 2E_A … (1)

But by Gauss’s Law

Φ = \(\frac{q}{\varepsilon_{0}}=\frac{\sigma A}{\varepsilon_{0}}\) [∵ q = σA] … (2)

From equations (1) and (2), we have

2E_A = \(\frac{\sigma A}{\varepsilon_{0}}\) … (3)

E = \(\frac{\sigma}{2 \varepsilon_{0}}\) …. (4)

This gives the electric field due to an infinite plane sheet of charge which is independent of the distance from the sheet.

(b) An infinitely large thin plane sheet has a uniform surface charge density +σ. Obtain the expression for the amount of work done in bringing a point charge q from infinity to a point, distant r, in front of the charged plane sheet.
Answer:
Work done

Numerical Problems:

Formulae for solving numerical problems

• q = ±ne
• F = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}\)
• Fmed = \(\frac{1}{4 \pi \varepsilon_{0} K} \frac{q_{1} q_{2}}{r^{2}}\) where K is dielectric constant.
• 
• E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{2 p}{r^{3}}\) for an electric dipole on its axial line.
• E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{p}{r^{3}}\) for an electric dipole on its equitorial line.
• Torque on an electric dipole in a uniform electric field, τ = PE sin θ.
• U = -pE cos θ, the potential energy of an electric dipole.
• Φ = \(\oint \vec{E} \cdot d \vec{A}=\frac{q_{\text {in }}}{\varepsilon_{0}}\)
• E = \(\frac{\sigma}{2 \varepsilon_{0}}\) , electric field due to an infinite plane sheet of charge.

Question 1.
An electric dipole of length 2 cm is placed with its axis making an angle of 60° to a uniform electric field of 105 NC^-1. If it experiences a torque of 8\(\sqrt{3}\) Nm, calculate the (i) magnitude of the charge on the dipole and (ii) potential energy of the dipole.
Answer:
(i) Using
q = τ / L E sin Φ

we have
q = 8N\(\sqrt{3}\) × 2 / 2 × 10^-2 × 10⁵ × \(\sqrt{3}\) = 8 × 10^-3 C

(ii) Using U = -pE cos θ, we have
U = – 8 × 10^-3 × 0.02 × 10⁵ × 0.5 = – 8 J

Question 2.
A charge of 17.7 × 10^-4 C is distributed uniformly over a large sheet of area 200 m². Calculate the electric field intensity at a distance of 20 cm from it in the air.
Answer:
Given q = 17.7 × 10^-4C, A = 200 m², r = 20cm = 0.2 m

Using the relation

Question 3.
Two fixed point charges +4e and +e units are separated by a distance ‘a’. Where should the third point charge be placed for it to be in equilibrium?
Answer:
The third charge will be in equilibrium if it experiences zero net force. Let it be placed at a distance x from the charge 4e.

4(a – x)² = x² solving for x we have
2 (a – x) = x or x = a/3

Question 4.
An electric field along the x-axis is given by \(\vec{E}\) = 100 îN/C for x > 0 and \(\vec{E}\) = -100 îN/C for x < 0. A right circular cylinder of length 20 cm and radius 5 cm lies parallel to the x-axis with its centre at the origin and one face at x = +10 cm, the other face at x = -10 cm. Calculate the net outward flux through the cylinder.

r = 5 cm = 0.05 m
Φ = ?

Net outward flux through the cylinder
Φ = Φ₁ + Φ₂ + Φ₃
= \(\vec{E} \cdot \overrightarrow{d s}_{1}+\vec{E} \cdot \overrightarrow{d s}_{2}+\vec{E} \cdot \overrightarrow{d s}_{3}\)

= Eds₁ cos 180° + Eds₂ cos 90° + Eds3 cos 0°
= – Eds₁ + Eds₂ (0) + Eds cos 0°
= – (-100) ds + 100 ds
= (100 +100) ds
= 200 × πr² = 200 × 3.14 × (0.05)²
= 1.57 Nm² C^-1

Question 5.
A hollow cylindrical box of length 1m and area of cross-section 25 cm² is placed in a three-dimensional coordinate system as shown in the figure. The electric field
in the region is given by \(\vec{E}\) = 50x î, where E is in NC^-1 and x is in meters. Find

(i) Net flux through the cylinder.
Answer:
We can see from the figure that on the left face E and dS are antiparallel. Therefore, the flux is
ΦL = \(\vec{E}\).\(\vec{dS}\) = E dS cos 180°
= -50 × 1 × 25 × 10^-4 = -0.125 N m² C^-1

On the right face, E and dS are parallel and therefore
ΦR = \(\vec{E}\).\(\vec{ds}\) = EdS cos 0°
= 50 × 2 × 25 × 10^-4 = 0.250 N m² C^-1

Therefore net flux is – 0.125 + 0.250 = + 0.125 N m² C^-1

(ii) Charge enclosed by the cylinder. (CBSE Delhi 2013)

Answer:
q = Φε₀ = 0.125 × 8.854 × 10^-12 = 1.1 × 10^-12C

Question 6.
(i) Define electric flux. Write its SI units,
Answer:
(i) It is defined as the number of electric field lines crossing a unit area perpendicular to the given area. It is measured in N m² C^-1

(ii) Consider a uniform electric field \(\vec{E}\) = 5 × 10³î NC^-1. Calculate the flux of this field through a square surface of area 12 cm² when
(a) its plane is parallel to the Y – Z plane, and
(b) the normal to its plane makes a 60° angle with the x-axis. (CBSE Delhi 2013C)
Answer:
Given \(\vec{E}\) = 5 × 10³î N C^-1, A = 12 cm² = 12 × 10^-4 m²

(a) Here \(\vec{E}\) = 5 × 103î N C^-1
Area of square = 12 × 10^-4 m²

The plane of surface area being parallel to YZ plane, hence
A = 12 × 10^-4 î m²

Electric flux of the field
Φ = \(\vec{E}\) \(\vec{A}\) = (5 × 10³ î). (12 × 10^-4î) = 6 N C^-1 m²
(b) When normal to the plane of surface area makes an angle of 60° with the X-axis, the flux is given by
Φ = EA cos 0 = 5 × 10³ × 12 × 10^-4 × 0.5
= 3 NC^-1 m²