NCERT Exemplar Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance give accurate information and knowledge in easytounderstand language. By solving the NCERT Physics exemplar problems available at BYJU’S, you will be able to grasp the topics covered in Chapter 2, Electrostatic Potential and Capacitance, thoroughly.
NCERT Class 12 Physics Exemplar Chapter 2 is an important study resource for the students as it consists of questions from the NCERT Exemplar Class 12 Physics book. NCERT Class 12 Physics Exemplar for Chapter 2 Electrostatic Potential and Capacitance PDF has extra questions prepared by our subject experts in different forms like fill in the blanks, match the following, short and long answer questions, true or false questions, etc.
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Multiple Choice Questions I
2.1. A capacitor of 4μF 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) 0
b) 4μC
c) 16μC
d) 8μC
Answer:
The correct answer is d) 8μC
2.2. A positively charged particle is released from rest in a 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
Answer:
The correct answer is c) decreases because the charge moves along the electric field
2.3. 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 the 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 fig (i)
Answer:
The correct answer is c) the work done is the same in fig (i), fig (ii), and fig (iii)
2.4. The electrostatic potential on the surface of a charged conducting sphere is 100V. Two statements are made in this regard:
S_{1}: At any point inside the sphere, the electric intensity is zero
S_{2}: At any point inside the sphere, the electrostatic potential is 100V
Which of the following is a correct statement?
a) S_{1} is true but S_{2} is false
b) Both S_{1} and S_{2} are false
c) S_{1} is true, S_{2} is also true, and S_{1} is the cause of S_{2}
d) S_{1} is true, and S_{2} is also true, but the statements are independent
Answer:
The correct answer is c) S_{1} is true, S_{2} is also true, and S_{1} is the cause of S_{2}
2.5. 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
Answer:
The correct answer is a) spheres
2.6. A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness d_{1} and dielectric constant k_{1,} and the other has thickness d_{2} and dielectric constant k_{2} as shown in the figure. This arrangement can be thought of as a dielectric slab of thickness d = d_{1} + d_{2} and effective dielectric constant k. The k is
a) k_{1}d_{1} + k_{2}d_{2}/d_{1}+d_{2}
b) k_{1}d_{1} + k_{2}d_{2}/k_{1} + k_{2}
c) k_{1}k_{2} (d_{1} + d_{2})/(k_{1}d_{1} + k_{2}d_{2})
d) 2k_{1}k_{2}/k_{1} + k_{2}
Answer:
The correct answer is c) k_{1}k_{2} (d_{1} + d_{2})/(k_{1}d_{1} + k_{2}d_{2})
Multiple Choice Questions II
2.7. 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 xy plane for a given z
Answer:
The correct answer is
b) for any x for a given z
c) for any y for a given z
d) on the xy plane for a given z
2.8. Equipotential surfaces
a) are closer in regions of large electric fields compared to regions of lower electric fields
b) will be more crowded near the sharp edges of a conductor
c) will be more crowded near regions of large charge densities
d) will always be equally spaced
Answer:
The correct answer is
a) are closer in regions of large electric fields compared to regions of lower electric fields
b) will be more crowded near the sharp edges of a conductor
c) will be more crowded near regions of large charge densities
2.9. 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 nonzero value
Answer:
The correct answer is
b) must be defined as
c) is zero
2.10. 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
Answer:
The correct answer is
b) the electric field is zero
c) there can be no charge inside the region
2.11. In the circuit shown in the figure, initially, key K_{1} is closed, and K_{2} is open. Then K_{1} is opened, and K_{2} is closed. Then
a) charge on C_{1} gets redistributed such that V_{1} = V_{2}
b) charge on C_{1} gets redistributed such that Q_{1}’ = Q_{2}’
c) charge on C_{1} gets redistributed such that C_{1}V_{1} + C_{2}V_{2} = C_{1}E
d) charge on C_{1} gets redistributed such that Q_{1}’ + Q_{2}’ = Q
Answer:
The correct answer is
a) charge on C_{1} gets redistributed such that V_{1} = V_{2}
d) charge on C_{1} gets redistributed such that Q_{1}’ + Q_{2}’ = Q
2.12. 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
Answer:
The correct answer is
a) there must be charges on the surface or inside itself
b) there cannot be any charge in the body of the conductor
2.13. A parallel plate capacitor is connected to a battery, as shown in the figure. Consider two situations:
A: Key K is kept closed, and plates of capacitors are moved apart using the insulating handle
B: Key K is opened, and plates of capacitors are moved apart using the insulating handle
Choose the correct options
a) In A: Q remains the same but C changes
b) In B: V remains the same, but C changes
c) In A: V remains the same, and hence Q changes
d) In B: Q remains the same, and hence V changes
Answer:
The correct answer is
c) In A: V remains the same, and hence Q changes
d) In B: Q remains the same, and hence V changes
Very Short Answers
2.14. Consider two conducting spheres of radii R_{1} and R_{2} with R_{1} > R_{2}. 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.
Answer:
There are two spheres
σ_{1}R_{1} = σ_{2}R_{2}
σ_{1}/ σ_{2} = R_{2}/R_{1}
As R_{2} > R_{1}, that means σ_{1} > σ_{2}
Therefore, the charge density of the smaller sphere is more than that of the larger sphere.
2.15. Do free electrons travel to a region of higher potential or lower potential?
Answer:
The force on the charged particle in the electric field is F = qE
The direction of the electric field and the direction of electrostatic force experienced by the free electrons are in opposite directions
The direction of the electric field is higher than the potential, and therefore, the electrons travel from a lower potential region to a higher potential.
2.16. Can there be a potential difference between two adjacent conductors carrying the same charge?
Answer:
Yes, there can be a potential difference between two adjacent conductors carrying the same charge.
2.17. Can the potential function have a maximum or minimum in free space?
Answer:
No, the potential function cannot be maximum or minimum in free space as the absence of atmosphere around the conductor prevents the electric discharge.
2.18. A test charge q is made to move in the electric field of a point charge Q along two different closed paths. The first path has sections along and perpendicular to lines of the electric field, and the second path is a rectangular loop of the same area as the first loop. How does the work done compare in the two cases?
Answer:
Work done is zero in both cases. The work done by the electric force on the charge is in the closed loop, which is equal to zero.
Short Answers
2.19. Prove that a closed equipotential surface with no charge within itself must enclose an equipotential volume.
Answer:
In a closed equipotential surface, the potential changes from position to position.
The potential inside the surface is different from the potential gradient caused in the surface that is dV/dr
This also means that the electric field is not equal to zero, and it is given as E = dV/dr
Therefore, it could be said that the field lines are either pointing inwards or outwards the surface.
So, it can be said that the field lines originate from the charges inside, which contradicts the original assumption. Therefore, the volume inside the surface must be equipotential.
2.20. 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, energy stored in it, electric field, the charge stored, and voltage will increase, decrease, or remain constant.
Answer:
Quantity  Battery is removed  Battery remains connected 
Capacity  C’ = KC  C’ = KC 
Charge  Q’ = Q  Q’ = KQ 
Potential  V’ = V/K  V’ V 
Intensity  E’ = E/K  E’ = E 
Energy  U’ = U/K  U’ = UK 
2.21. Prove that if an insulated, uncharged conductor is placed near a charged conductor and no other conductors are present, the uncharged body must be intermediate in potential between that of the charged body and that of infinity.
Answer:
The electric potential decreasing along the direction of the electric field is given as E = dV/dr
The electric potential decreases when the path from the charged conductor is taken to the uncharged conductor along the direction of the electric field.
This continues when another uncharged conductor is considered to the infinity lowering the potential even further.
This shows that the uncharged body is at intermediate potential, and the charged body is at infinity potential.
2.22. Calculate the potential energy of a point charge –q placed along the axis due to 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 the graph, can you see what would happen if –q is displaced slightly from the centre of the ring?
Answer:
U is the potential energy of a point charge q which is placed at potential V, U = qV
A negatively charged particle is placed at the axis of the ring with charge Q
Let a be the radius of the ring
The electric potential at the axial distance is given as
The potential energy, U is given as
When z = infinity, U = 0.
When z = 0, U is given as
2.23. Calculate potential on the axis of a ring due to charge Q uniformly distributed along the ring of radius R.
Answer:
Point P is considered to be at a distance z from the centre of the ring. The charge dq is at a distance z from point P.
V is written as
Therefore, the net potential is given as
Long Answers
2.24. Find the equation of the equipotential for an infinite cylinder of radius r_{0}, carrying charge of linear density λ.
Answer:
The equation of the equipotential for an infinite cylinder of radius r_{0} with linear charge density λ is given as:
r = r_{0}e 2πε_{0}/ λ [V(r) – V(r_{0})]
2.25. Twopoint charges of magnitude +q and –q are placed at (d/2, 0, 0) and (d/2, 0, 0), respectively. Find the equation of the equipotential surface where the potential is zero.
Answer:
The potential at point P due to charges is given as
Vp = 1/4πε_{0} q/r_{1} + 1/4 πε_{0} (q)/r_{2}
The net electric potential at this point is zero,
Therefore, r_{1} = r_{2}
We know that
To solve the two equations, we require the equation inplane x = 0, which is a yz plane.
2.26. A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage (U) as ε = αU where α = 2V^{1}. A similar capacitor with no dielectric is charged to U_{0} = 78 V. It is then connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors.
Answer:
Since the capacitors are connected in parallel, the potential difference across the capacitors is the same. The final voltage is assumed to be U. C is the capacitance of the capacitor without dielectric, then the charge is given as Q_{1} = CU
The initial charge is given as
Q_{0} = CU_{0}
The conversion of charges is
Q_{0} = Q_{1} + Q_{2}
CU_{0} = CU + αCU_{2}
αU_{2} + U – U_{0} = 0
Solving the equation, we get U = 6V
2.27. 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 the centre of the bottom plate. Find the minimum voltage required to lift the disc if the mass of the disc is m.
Answer:
When the conducting disc is placed at the centre of the bottom plate, the potential of the disc is equal to the potential of the plate. The electric field on the disc is given as
E = V/d
The charge q’ is transferred to the disc, which is given as
q’ = ε_{0} V/d πr^{2}
The force acting on the disc is
F = ε_{0} V^{2}/d^{2} πr^{2}
Therefore, V = square root of mdg/ π ε_{0}r^{2}
2.28. a) In a quark model of elementary particles, a neutron is made of one up quark and two down quarks. Assume that they have a triangle configuration with a side length of the order of 10^{15} m. Calculate the electrostatic potential energy of the neutron and compare it with its mass 939 MeV.
b) Repeat the above exercise for a proton which is made of two up and one down quarks.
Answer:
There are three charges in the system. The potential energy of the system is equal to the sum of the PE of each pair.
U = 1/4 π ε_{0} {q_{d}q_{d}/r – q_{u}q_{d}/r – q_{u}q_{d}/r}
Substituting the values we get,
U = 5.11 10^{4}
2.29. Two metal spheres, one of radius R and the other of radius 2R, both have the same surface charge density σ. They are brought in contact and separated. What will be the new surface charge densities on them?
Answer:
Following are the charges on the metal sphere before contact
Q_{1} = σ.4πR^{2}
Q_{2} = σ.4π(2R)^{2} = 4Q_{1}
Q_{1}’ and Q_{2}’ are the charges on the metal sphere after contact
Q_{1}’ + Q_{2}’ = Q_{1} + Q_{2} = 5Q_{1}
When the metal spheres are in contact, the following is the potentials acquired by them
Q_{1}’ = Q_{2}’/2
Solving the equations we
σ_{1} = 5 σ/3
σ_{2} = 5 σ/6
2.30. In the circuit shown in the figure, initially, K_{1} is closed, and K_{2} is open. What are the charges on each capacitor? Then K_{1} was opened, and K_{2} was closed. What will be the charge on each capacitor now?
Answer:
When K_{1} is closed, and K_{2} is open, the capacitors C_{1} and C_{2} are connected in series with the battery
Therefore, the charge in capacitors C_{1} and C_{2} are
Q_{1} = Q_{2} = q = (C_{1}C_{2}/C_{1}+C_{2})E = 18μC
When the capacitors C_{2} and C_{3} are placed in parallel,
C_{2}V’ + C_{3}V’ = Q_{2}
V’ = Q_{2}/C_{2} + C_{3} = 3V
Therefore,
Q_{2}’ = 3CV’ = 9 μC
Q_{3}’ = 3CV’ = μC
Q_{1}’ = 18 μC
2.31. Calculate potential on the axis of a disc of radius R due to a charge Q uniformly distributed on its surface.
Answer:
In the above figure, we can see that the disc is divided into several charged rings. Let P be the point on the axis of the disc at a distance x from the centre of the disc.
The radius of the ring is r and the width is dr. dq is the charge on the ring, which is given as
dq = σdA = σ2πrdr
The potential is given as
The total potential at P is given as
Q/2πε_{0}R^{2} (√R^{2} + x^{2} –x)
2.32. Two charges, q_{1} and q_{2,} are placed at (0, 0, d) and (0, 0, d), respectively. Find the locus of points where the potential is zero.
Answer:
We know that the potential at point P is V = ∑V_{i}
Where V_{i} = q_{i}/4πε_{0}, r_{i} is the magnitude of the position vector P
V = 1/4 πε_{0} ∑ q_{i}/r_{pi}
When the (x,y,z) plane is considered, the two charges lie on the zaxis and are separated by 2d. The potential is given as
Squaring the equation, we get
x^{2} + y^{2} + z^{2} + [(q_{1}/q_{2})^{2}+1/(q_{1}/q_{2})^{2}1] (2zd)+d^{2}0
Therefore, the equation of the sphere is
x^{2} + y^{2} + z^{2} + 2ux + 2uy + 2wz + g = 0
The centre of the sphere is
And radius is
2.33. Two charges –q each are separated by distance 2d. A third charge +q is kept at midpoint O. Find potential energy of +q as a function of small distance x from O due to –q charges. Sketch PE versus x and convince yourself that the charge at O is in an unstable equilibrium.
Answer:
In the above figure, +q is the charge that got displaced from O towards (d,0).
This is written as
At x = 0
Differentiating the equation with respect to x, we get
When x < 0, dU/dx > 0
And when x > 0, dU/dx < 0
Using this, we can define the charge on the particle to be F = dU/dx
F = dU/dx = 0
When
d^{2}U/dx^{2} = positive, equilibrium is stable
d^{2}U/dx^{2} = negative, equilibrium is unstable
d^{2}U/dx^{2} = 0, equilibrium is neutral
Therefore, when x = 0, d^{2}U/dx^{2} = (2dq^{2}/4πε0)(1/d^{6})(2d^{2}) < 0
Which shows that the system is in unstable equilibrium.
Electric charges which are slow or stationary is called Electrostatic Charge, and the amount of work needed to move from one initial point to another is termed Electrostatic Potential. It can be measured in volts and is a scalar quantity. Capacitance, on the other hand, can be defined as the ability of a capacitor to store energy. They are measured in farads.
Important Concepts Covered in NCERT Exemplar Electrostatic Potential and Capacitance
Sl.No.  Concept Name 
1  Electrostatic Potential – derivation of the electric formula 
2  Potential Due to a Point Charge – relation between electric charge and potential 
3  Potential Due to an Electric Dipole – derivation of potential access formula 
4  Potential Due to a System of Charges – derivation of the potential energy of a system of two charges 
5  Equipotential Surfaces – derivation of potential due to charge 
6  The Relation between Field and Potential:

7  Potential Energy of a System of Two Charges in an External Field 
8  Potential Energy of a Dipole in an External Field 
9  Electrostatics of Conductors in Different Conditions, Electroshielding 
10  Dielectrics and Polarisation – meaning dielectric constant derivation 
11  Capacitors and Capacitance – meaning, capacitance derivation and problems 
12  The ParallelPlate Capacitor – meaning and derivation 
13  Effect of Dielectric on Capacitance – derivations and numerical problems 
14  Combination of Capacitors – capacitors in series, capacitors in parallel 
15  Energy Stored in a Capacitor – Energy density of electric field, derivation 
16  Van De Graaff Generator – introduction, applications 
Also Access 
NCERT Solutions for Class 12 Physics Chapter 2 
CBSE Notes for Class 12 Physics Chapter 2 
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