Multiple Choice Questions (MCQs)

Q. 1 A capacitor of 4 µF is connected as shown in the circuit. The internal resistance of the battery is 0.5Ω. The amount of charge on the capacitor plates will be

(a) O  (b) 4 µ C  (c) 16 µC  (d) 8 µC

Ans.  (d) Current flows through 2Ω resistance from left to right, is given by

I = V =   = 1A

The potential difference across 2Ω resistance V = IR = 1 x 2 = 2V

Since, capacitor is in parallel with 2Q resistance, so it also has 2V potential difference across it.

The charge on capacitor

q = CV = (2µF) x 2V = 8µC

Note The potential difference across 2Ω resistance solely occurs across capacitor as no potential drop occurs across 10Ω resistance.

Q. 2 A positively charged particle is released from rest in an uniform electric field. The electric potential energy of the charge

(a) remains a constant because the electric field is uniform

(b) increases because the charge moves along the electric field

(c) decreases because the charge moves along the electric field

(d) decreases because the charge moves opposite to the electric field Thinking Process

In this problem, the relationship between £ and V is actualised.

Ans.  (c) The direction of electric field is always perpendicular to one equipotential surface maintained at high electrostatic potential to other equipotential surface maintained at low electrostatic potential.

The positively charged particle experiences electrostatic force along the direction of electric field i.e., from high electrostatic potential to low electrostatic potential. Thus, the work is done by the electric field on the positive charge, hence electrostatic potential energy of the positive charge decreases.

Q. 3 Figure shows some equipotential lines distributed in space. A charged object is moved from point A to point B.

(a) The work done in Fig. (i) is the greatest

(b) The work done in Fig. (ii) is least

(c) The work done is the same in Fig. (i), Fig.(ii) and Fig. (iii)

(d) The work done in Fig. (iii) is greater than Fig. (ii) but equal to that in

Ans.  (c) The work done by a electrostatic force is given by W₁₂ = q (V₂ – V₁). Here initial and final potentials are same in all three cases and same charge is moved, so work done is same in all three cases.

Q. 4 The electrostatic potential on the surface of a charged conducting sphere is 100V. Two statements are made in this regard

S₁ at any point inside the sphere, electric intensity is zero.

S₂ at any point inside the sphere, the electrostatic potential is 100V.

Which of the following is a correct statement?

(a) S¹ is true but S₂ is false

(b) Both S₁ and S₂ are false

(c) S¹ is true, S₂ is also true and S¹ is the cause of S₂

(d) S¹ is true, S₂ is also true but the statements are independant

Ans.  (c) In this problem, the electric field intensity E and electric potential V are related as

E = – 

Electric field intensity E = 0 suggest that  = 0

This imply that V = constant.

Thus, E = 0 inside the charged conducting sphere causes , the same electrostatic potential 100V at any point inside the sphere.

Note V equals zero does not necessary imply that E = 0 e.g., the electric potential at any point on the perpendicular bisector due to electric dipole is zero but E not.

E = 0 does not necessary imply that V = 0 e.g., the electric field intensity at any point inside the charged spherical shell is zero but there may exist non–zero electric potential.

Q. 5 Equipotential at a great distance from a collection of charges whose total sum is not zero are approximately

(a) spheres  (b) planes

(c) paraboloids  (d) ellipsoids

Ans.  (a) In this problem, the collection of charges, whose total sum is not zero, with regard to great distance can be considered as a point charge. The equipotential due to point charge are spherical in shape as electric potential due to point charge q is given by

V = K_e

This suggest that electric potentials due to point charge is same for all equidistant points. The locus of these equidistant points, which are at same potential, form spherical surface.

Q. 6 A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness d₁ and dielectric constant K₁ and the other has thickness d₂ and dielectric constant K₂ as shown in figure. This arrangement can be thought as a dielectric slab of thickness d ( = d₁ + d₂) and effective dielectric constant K. The K is’

(a)  (b)

(c)  (d)

Ans.  (c) The capacitance of parallel plate capacitor filled with dielectric block has thickness c₁ and dielectric constant K₂ is given by

C₁ = 

Similarly, capacitance of parallel plate capacitor filled with dielectric block has thickness d₂ and dielectric constant K₂ is given by

C₂ = 

Since, the two capacitors are in series combination, the equivalent capacitance is given by

But the equivalent capacitances is given by

C = 

On comparing, we have

K = 

Note For the equivalent capacitance of the combination, thickness is equal to the separation between two plates i.e., d₁+ d₂ and dielectric constant K.

Multiple Choice Questions (More Than One Options)

Q. 7 Consider a uniform electric field in the (??)–direction. The potential is a constant

(a) in all space  (b) for any x for a given z

(c) for any y for a given z  (d) on the x–y plane for a given z

Ans.  (b, c, d)

Here, the figure electric field is always remain in the direction in which the potential decreases steepest. Its magnitude is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface at the point.

The electric field in z–direction suggest that equipotential surfaces are in x–y plane. Therefore the potential is a constant for any x for a given z, for any y for a given z and on the x–y plane for a given z.

Note The shape of equipotential surfaces depends on the nature and type of distribution of charge e.g., point charge leads to produce spherical surfaces whereas line charge distribution produces cylindrical equipotential surfaces.

Q. 8 Equipotential surfaces

(a) are closer in regions of large electric fields compared to regions of lower electric fields

(b) will be more crowded near sharp edges of a conductor

(c) will be more crowded near regions of large charge densities

(d) will always be equally spaced

Ans.  (a,b,c)

The electric field intensity E is inversely proportional to the separation between equipotential surfaces So, equipotential surfaces are closer in regions of large electric fields.

Since, the electric field intensities is large near sharp edges of charged conductor and near regions of large charge densities. Therefore, equipotential surfaces are closer at such places.

Q. 9 The work done to move a charge along an equipotential from A to B

(a) cannot be defined as – 

(b) must be defined as – 

(c) is zero

(d) can have a non–zero value

Ans.  (c) Work done in displacing a charge particle is given by W₁₂ = q(V₂ – V¹) and the line integral of electrical field from point 1 to 2 gives potential difference V₂ – V₁ = –For equipotential surface, V₂ – V₁ = O and W = 0.

Note: If displaced charged particle is + 1 C, then and only then option (b) is correct. But the NCERT exemplar book has given (b) as correct options which probably not so under given conditions.

Q. 10 In a region of constant potential

(a) the electric field is uniform

(b) the electric field is zero

(c) there can be no charge inside the region

(d) the electric field shall necessarily change if a charge is placed outside the region

Ans.  (b, c)

The electric field intensity E and electric potential V are related as E = 0 and for V = constant,  = 0

This imply that electric field intensity E = 0.

Q. 11 In the circuit shown in figure initially key K₁ is closed and key K₂ is open. Then K₁ is opened and K₂ is closed (order is important).

[Take Q’₁ and Q₂‘ as charges on C₁ and C₂ and V₁ and V₂ as voltage respectively.]

Then,

(a) charge on C₁ gets redistributed such that V₁ = V₂

(b) charge on C₁ gets redistributed such that Q’₁ = Q’₂

(c) charge on C¹ gets redistributed such that C₁V₁ + C₂V₂ = C₁E

(d) charge on C, gets redistributed such that Q’₁ + Q’₂ = Q

Ans.  (a, d)

The charge stored by capacitor C₁ gets redistributed between C₁ and C₂ till their potentials become same i.e. ,V₂ = V₁ By law of conservation of charge, the charge stored in capacitor C₁ when key K₁ is closed and key K₂ is open is equal to sum of charges on capacitors C₁ and C₂ when K₁ is opened and K₂ is closed i.e.,

Q₁ + Q₂ = Q

Q. 12 If a conductor has a potential V ≠ 0 and there are no charges anywhere else outside, then

(a) there must be charges on the surface or inside itself

(b) there cannot be any charge in the body of the conductor

(c) there must be charges only on the surface

(d) there must be charges inside the surface

Ans.  (a, b)

The charge resides on the outer surface of a closed charged conductor.

Q. 13 A parallel plate capacitor is connected to a battery as shown in figure. Consider two situations.

A. Key K is kept closed and plates of capacitors are moved apart using insulating handle.

B. Key K is opened and plates of capacitors are moved apart using insulating handle.

Choose the correct option(s).

(a) In A Q remains same but C changes

(b) In B V remains same but C changes

(c) In A V remains same and hence Q changes

(d) In B Q remains same and hence V changes

Ans.  (c, d)

Case A When key K is kept closed and plates of capacitors are moved apart using insulating handle, the separation between two plates increases which in turn decreases its capacitance  and hence, the charge stored decreases as Q = CV ( potential continue to be the same as capacitor is still connected with cell).

Case B When key K is opened and plates of capacitors are moved apart using insulating handle, charge stored by disconnected charged capacitor remains conserved and with the decreases of capacitance, potential difference V increases as V = Q /C.

Very Short Answer Type Questions

Q. 14 Consider two conducting spheres of radii R₁ and R₂ with R₁ > R₂. If the two are at the same potential, the larger sphere has more charge than the smaller sphere. State whether the charge density of the smaller sphere is more or less than that of the larger one.

Ans.  Since, the two spheres are at the same potential, therefore

= 

This imply that σ₁ > σ₂.

The charge density of the smaller sphere is more than that of the larger one.

Q. 15 Do free electrons travel to region of higher potential or lower potential?

Ans.  The free electrons experiences electrostatic force in a direction opposite to the direction of electric field being is of negative charge. The electric field always directed from higher potential to lower travel.

Therefore, electrostatic force and hence direction of travel of electrons is from lower potential to region of higher potential.

Q. 16 Can there be a potential difference between two adjacent conductors carrying the same charge?

Ans.  Yes, if the sizes are different.

Q.17  Can the potential function have a maximum or minimum in free space?

Ans.  No, The absence of atmosphere around conductor prevents the phenomenon of electric discharge or potential leakage and hence, potential function do not have a maximum or minimum in free space.

Q. 18 A test charge q is made to move in the electric field of a point charge Q along two different closed paths [figure first path has sections along and perpendicular to lines of electric field. Second path is a rectangular loop of the same area as the first loop. How does the work done compare in the two cases?

Ans.  As electric field is conservative, work done will be zero in both the cases.

Note Conservative forces (like electrostatic force or gravitational force) are those forces, work done by which depends only on initial position and final position of object viz charge, but not on the path through which it goes from initial position to final position.

Short Answer Type Questions

Q. 19 Prove that a closed equipotential surface with no charge within itself must enclose an equipotential volume.

Ans.  Let’s assume contradicting statement that the potential is not same inside the closed equipotential surface. Let the potential just inside the surface is different to that of the surface causing in a potential gradient. Consequently electric field comes into existence, which is given by as E = – dV.

Consequently field lines pointing inwards or outwards from the surface. These lines cannot be again on the surface, as the surface is equipotential. It is possible only when the other end of the field lines are originated from the charges inside.

This contradict the original assumption. Hence, the entire volume inside must be equipotential.

Q. 20 A capacitor has some dielectric between its plates and the capacitor is connected to a DC source. The battery is now disconnected and then the dielectric is removed . State whether the capacitance, the energy stored in it, electric field, charge stored and the voltage will increase, decrease or remain constant.

Ans.  The capacitance of the parallel plate capacitor, filled with dielectric medium of dielectric constant K is given by

C = , where signs are as usual.

The capacitance of the parallel plate capacitor decreases with the removal of dielectric medium as for air or vacuum K = 1.

After disconnection from battery charge stored will remain the same due to conservation of charge.

The energy stored in an isolated charge capacitor = ; as q is constant, energy stored ∞ 1/C and C decreases with the removal of dielectric medium, therefore energy stored increases. Since q is constant and V = q / C and C decreases which in turn increases V and therefore £ increases as E = V/d .

Note: One of the very important questions with the competitive point of view.

Q. 21 Prove that, if an insulated, uncharged conductor is placed near a charged conductor and no other conductors are present, the uncharged body must intermediate in potential between that of the charged body and that of infinity.

Ans.  Let us take any path from the charged conductor to the uncharged conductor along the direction of electric field. Therefore, the electric potential decrease along this path.

Now, another path from the uncharged conductor to infinity will again continually lower the potential further. This ensures that the uncharged body must be intermediate in potential between that of the charged body and that of infinity.

Q. 22 Calculate potential energy of a point charge –q placed along the axis due to a charge+ Q uniformly distributed along a ring of radius R. Sketch PE, as a function of axial distance z from the centre of the ring. Looking at graph, can you see what would happen if –q is displaced slightly from the centre of the ring (along the axis)?

Ans.  Let us take point P to be at a distance x from the centre of the ring, as shown in figure. The charge element dq is at a distance x from point P. Therefore, V can be written as

where, k = , since each element dq is at the same distance from point P, so we have net potential

Considering – q charge at P, the potential energy is given by

U = W = q x potential difference

This is the required expression.

The variation of potential energy with z is shown in the figure. The charge –q displaced would perform oscillations.

Nothing can be concluded just by looking at the graph.

Q. 23 Calculate potential on the axis of a ring due to charge Q uniformly distributed along the ring of radius R.

Ans.  Let us take point P to be at a distance x from the centre of the ring, as shown in figure. The charge element dq is at a distance x from point P. Therefore, V can be written as

where, k_e = , since each element dq is at the same distance from point P, so we have net potential

Long Answer Type Questions

Q. 24 Find the equation of the equipotentials for an infinite cylinder of radius r₀ carrying charge of linear density .

Ans.  Let the field lines must be radically outward. Draw a cylindrical Gaussian surface of radius r and length l. Then, applying Gauss’ theorem

The equipotential surfaces are cylinders of radius.

Q. 25 Two point charges of magnitude + q and – q are placed at (– d / 2, 0, 0) and (d / 2, 2, 0), respectively. Find the equation of the equipotential surface where the potential is zero.

Ans.  Let the required plane lies at a distance x from the origin as shown in figure.

Q. 26 A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage (U) as ε = αU where a = 2V^–1. A similar capacitor with no dielectric is charged to U₀ = 78 V. It is then connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors.

Ans.  Assuming the required final voltage be U. If C is the capacitance of the capacitor without the dielectric, then the charge on the capacitor is given by Q₁ = CU

Since, the capacitor with the dielectric has a capacitance εC. Hence, the charge on the capacitor is given by

Q. 27 A capacitor is made of two circular plates of radius R each, separated by a distance d << R. The capacitor is connected to a constant voltage. A thin conducting disc of radius r << R and thickness t << r is placed at a centre of the bottom plate. Find the minimum voltage required to lift the disc if the mass of the disc is m.

Ans.  Assuming initially the disc is in touch with the bottom plate, so the entire plate is a equipotential.

The electric field on the disc, when potential difference V is applied across it, given by

Let charge q is transferred to the disc during the process,

Therefore by Gauss’ theorem,

Q. 28 (a) In a quark model of elementary particles, a neutron is made of one up quarks [charge (2/3) e] and two down quarks [charges –(1/3) e]. Assume that they have a triangle configuration with side length of the order of 10^–15 m. Calculate electrostatic potential energy of neutron and compare it with its mass 939 MeV.

(b) Repeat above exercise for a proton which is made of two up and one down quark.

Ans.  This system is made up of three charges. The potential energy of the system is equal to the algebraic sum of PE of each pair. So,

Q. 29 Two metal spheres, one of radius R and the other of radius 2R, both have same surface charge density σ. They are brought in contact and separated. What will be new surface charge densities on them?

Ans.  The charges on two metal spheres, before coming in contact, are given by

Q = σ.4R²

Q₂ = σ.4(2R²)

= 4(σ.4R²) = 4Q₁

Let the charges on two metal spheres , after coming in contact becomes O’₁ and Q’₂.

Now applying law of conservation of charges is given by

After coming in contract, they acquire equal potentials. Therefore, we have

Q. 30 In the circuit shown in figure, initially K₁ is closed and K₂ is open. What are the charges on each capacitors?

Then K₁ was opened and K₂ was closed (order is important), what will be the charge on each capacitor now? [C = 1µF]

Ans.  In the circuit, when initially K₁ is closed and K₂ is open, the capacitors C₁ and C₂ acquires potential difference V₁ and V₂ respectively. So, we have

V_i1 V₂ = E

and  V₁ + V₂ = 9V

Also, in series combination,  V  1 /c

V₁ : V₂ = 1/6:1/3

On solving

  V₁, = 3V and V₂ = 6V

 Q₁ = C₁V₁ = 6Cx 3 = 18µC

Q₂ = 9µC and Q₃ = 0

Then, K, was opened and K₂ was closed, the parallel combination of C₂ and C₃ is in series with C₁.

Q₂ = Q’₂+ Q₃

and considering common potential of parallel combination as V, then we have

C₂V+C₃V = Q₂

  V =  = (3/2)V

On solving,  Q’₂ = (9/2)µC

and  Q₃ = (9/2)µC

Q. 31 Calculate potential on the axis of a disc of radius R due to a charge Q uniformly distributed on its surface.

Ans.  Let the point P lies at a distance x from the centre of the disk and take the plane of the disk to be perpendicular to the x–axis. Let the disc is divided into a number of charged rings as shown in figure.

The electric potential of each ring, of radius r and width dr, have charge dq is given by

σ dA = σ2 πrdr

and potential is given by

(Refer the solution of Q. 23 )

where Kε = 1/4πε₀ the total electric potential at P, is given by

Note You may take a = R in this problem.

Q. 32 Two charges q₁ and q₂ are placed at (0, 0, d) and (0, 0, –d) respectively. Find locus of points where the potential is zero.

Ans.  Let any arbitrary point on the required plane is (x, y, z) .The two charges lies on z–axis at a separation of 2d.

The potential at the point P due to two charges is given by

Note The centre and radius of sphere (x–a)² +(y–b)² +(z–c)² = r² is (a, b, c) and r respectively.

Q. 33 Two charges –q each are separated by distance 2d. A third charge + q is kept at mid–point 0. Find potential energy of + q as a function of small distance x from 0 due to –q charges. Sketch PE Vs/x and convince yourself that the charge at 0 is in an unstable equilibrium.

Ans.  Let third charge + q is slightly displaced from mean position towards first charge. So, the total potential energy of the system is given by

The system will be in equilibrium, if

F =  = 0

On solving, x = 0. So for, +q charge to be in stable/unstable equilibrium, finding second derivative of PE.

This shows that system will be unstable equilibrium.

Ncert Exemplar problems with Solution Chapter 1 Electric Charge and Fields

Ncert Exemplar problems with Solutions Chapter 3 Current Electricity