CBSE Class 12 Physics Chapter-2 Important Questions – Free PDF Download
Free PDF download of Important Questions with Answers for CBSE Class 12 Physics Chapter 2 – Electrostatic Potential and Capacitance 3 Marks Questions prepared by expert Physics teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.
CBSE Class 12 Physics Chapter-2 Important Questions
CBSE Class 12 Physics Important Questions Chapter 2 – Electrostatic Potential and Capacitance
3 Marks Questions
1.Two dielectric slabs of dielectric constant 
are filled in between the two plates, each of area A, of the parallel plate capacitor as shown in the figure. Find the net capacitance of the capacitor? Area of each plate =
Ans. Here the two capacitors are in parallel
Net capacitance 
2. Prove that the energy stored in a parallel plate capacitor is given by
?
Ans. Suppose a capacitor is connected to a battery and it supplies small amount of change dq at constant potential V, then small amount of work done by the battery is given by
dw = Vdq
dw = qc/dq (Since q = CV)
Total work done where capacitor is fully changed to q.
This work done is stored in the capacitor in the form of electrostatic potential energy.
3. State Gauss’s Theorem in electrostatics? Using this theorem define an expression for the field intensity due to an infinite plane sheet of change of charge density 
?
Ans. Gauss’s Theorem states that electric flux through a closed surface enclosing a charge q in vacuum is 
times the magnitude of the charge enclosed
Is 
Consider a charge is distributed over an infinite sheet of area S having surface change density .
To enclose the charge on sheet an imaginary Gaussian surface cylindrical in shape is assumed and it is divided into three sections 
According to Gauss’s theorem
We know
For the given surface 
Combined eq. (1) & (2) 
E = 
E.2S = 
4. Derive an expression for the total work done in rotating an electric dipole through an angle 
in a uniform electric field?
Ans. We know 
= PE sin
If an electric dipole is rotated through an angled against the torque acting on it, then small amount of work done is
dw = d = PE sin d
For rotating through on angle , from 
w = – PE cos
5. If 
and
, calculate the equivalent capacitance of the given network between points A & B?
Ans. Since 
are in series
and C are in series
Here 
are in series
Cnet = 60/61 pF
6. Prove that energy stored per unit volume in a capacitor is given by 
, where E is the electric field of the capacitor?
Ans. We know capacitance of a parallel plate capacitor, 
, electric filed in between the plates 
where 
is the surface charge density of the plates.
Energy stored per unit volume = 
Energy stored per unit volume = 
(Volume of the capacitor = Ad)
Energy stored/volume = Hence proved
7. Keeping the voltage of the charging source constant. What would be the percentage change in the energy stored in a parallel plate capacitor if the separation between its plates were to be decreased by 10%?
Ans.
For parallel plate 
When d’ = d – 10% of d= 0.9 d
Then
Change in energy = U’-U = 
% change =
% change = 11.1%
8. Two identical plane metallic surfaces A and B are kept parallel to each other in air separated by a distance of 1.0 cm as shown in the figure. Surface A is given a positive potential of 10V and the outer surface of B is earthed.
(a) What is the magnitude and direction of uniform electric field between point Y and Z? What is the work done in moving a change of 20
from point X to Y?
(b) Can we have non-zero electric potential in the space, where electric field strength is zero?
Ans. (a)Since 
(ii) Since surface A is an equipotential surface ie 
Work done from X to Y = Zero .
(b) 
Or V= constant (non zero)
So we can have non-zero electric potential, where electric field is zero.
9. A regular hexagon of side 10 cm has a charge 5 
at each of its vertices. Calculate the potential at the centre of the hexagon.
Ans. The given figure shows six equal amount of charges, q, at the vertices of a regular hexagon.
Where,
Charge, 
Side of the hexagon, l = AB = BC = CD = DE = EF = FA = 10 cm
Distance of each vertex from centre O, d = 10 cm
Electric potential at point O,
Where,
= Permittivity of free space
Therefore, the potential at the centre of the hexagon is
.
10. A spherical conductor of radius 12 cm has a charge of 
distributed uniformly on its surface. What is the electric field
1. Inside the sphere
2. Just outside the sphere
3. At a point 18 cm from the centre of the sphere?
Ans. Radius of the spherical conductor, r = 12 cm = 0.12 m
Charge is uniformly distributed over the conductor, 
Electric field inside a spherical conductor is zero. This is because if there is field inside the conductor, then charges will move to neutralize it.
1. Electric field E just outside the conductor is given by the relation,
Where,
= Permittivity of free space
Therefore, the electric field just outside the sphere is 
.
2. Electric field at a point 18 m from the centre of the sphere = 
Distance of the point from the centre, d = 18 cm = 0.18 m
Therefore, the electric field at a point 18 cm from the centre of the sphere is 
.
11. Three capacitors each of capacitance 9 pF are connected in series.
1. What is the total capacitance of the combination?
2. What is the potential difference across each capacitor if the combination is connected to a 120 V supply?
Ans.Capacitance of each of the three capacitors, C = 9 pF
Equivalent capacitance 
of the combination of the capacitors is given by the relation,
Therefore, total capacitance of the combination is 
.
1. Supply voltage, V = 100 V
Potential difference 
across each capacitor is equal to one-third of the supply voltage.
Therefore, the potential difference across each capacitor is 40 V.
12. Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel.
1. What is the total capacitance of the combination?
2. Determine the charge on each capacitor if the combination is connected to a 100 V supply.
Ans. Capacitances of the given capacitors are
For the parallel combination of the capacitors, equivalent capacitor
is given by the algebraic sum,
Therefore, total capacitance of the combination is 9 pF.
1. Supply voltage, V = 100 V
The voltage through all the three capacitors is same = V = 100 V
Charge on a capacitor of capacitance C and potential difference V is given by the relation,
For C = 2 pF,
For C = 3 pF,
For C = 4 pF,
13. In a parallel plate capacitor with air between the plates, each plate has an area of 
and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?
Ans. Area of each plate of the parallel plate capacitor, A 
Distance between the plates, d = 3 mm 
Supply voltage, V = 100 V
Capacitance C of a parallel plate capacitor is given by,
Where,
= Permittivity of free space
Potential V is related with the charge q and capacitance C as
Therefore, capacitance of the capacitor is 17.71 pF and charge on each plate is 
.
14. Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates,
1. While the voltage supply remained connected.
2. After the supply was disconnected.
Ans.Dielectric constant of the mica sheet, k = 6
1. Initial capacitance, 
New capaci tan ce, 
Supply voltage, V = 100 V
New charge 
s
Potential across the plates remains 100 V.
2. Dielectric constant, k = 6
Initial capacitance
New capacitance 
If supply voltage is removed, then there will be no effect on the amount of charge in the plates.
Charge =
Potential across the plates is given by,
15. A 12 pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?
Ans. Capacitor of the capacitance, 
Potential difference, V = 50 V
Electrostatic energy stored in the capacitor is given by the relation,
Therefore, the electrostatic energy stored in the capacitor is 
16. A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?
Ans. Capacitance of the capacitor, C = 600 pF Potential difference, V = 200 V
Electrostatic energy stored in the capacitor is given by,
If supply is disconnected from the capacitor and another capacitor of capacitance C = 600 pF is connected to it, then equivalent capacitance of the combination is given by,
New electrostatic energy can be calculated as
Loss in electrostatic energy =
Therefore, the electrostatic energy lost in the process is
.
17. A spherical conducting shell of inner radius 
and outer radius 
has a charge
1. A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell?
2. Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.
Ans. (a) Charge placed at the centre of a shell is +q. Hence, a charge of magnitude –q will be induced to the inner surface of the shell. Therefore, total charge on the inner surface of the shell is –q.
Surface charge density at the inner surface of the shell is given by the relation,
A charge of +q is induced on the outer surface of the shell. A charge of magnitude Q is placed on the outer surface of the shell. Therefore, total charge on the outer surface of the shell is Q + q. Surface charge density at the outer surface of the shell,
1. Yes
The electric field intensity inside a cavity is zero, even if the shell is not spherical and has any irregular shape. Take a closed loop such that a part of it is inside the cavity
along a field line while the rest is inside the conductor. Net work done by the field in carrying a test charge over a closed loop is zero because the field inside the conductor is zero. Hence, electric field is zero, whatever is the shape.
18. If one of the two electrons of a
molecule is removed, we get a hydrogen molecular ion
. In the ground state of an, the two protons are separated by roughly 1.5 
, and the electron is roughly 1 from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.
Ans.
The system of two protons and one electron is represented in the given figure.
Charge on proton 1, 
Charge on proton 2, 
Charge on electron, 
Distance between protons 1 and 2, 
Distance between proton 1 and electron, 
Distance between proton 2 and electron, 
The potential energy at infinity is zero.
Potential energy of the system,
Substituting 
Therefore, the potential energy of the system is
.
19. Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.
Ans.Let a be the radius of a sphere A, QA be the charge on the sphere, and CA be the capacitance of the sphere. Let b be the radius of a sphere B, QB be the charge on the sphere, and CB be the capacitance of the sphere. Since the two spheres are connected with a wire, their potential (V) will become equal.
Let EAbe the electric field of sphere A and EB be the electric field of sphere B. Therefore, their ratio,
However 
And 
Putting the value of (2) in (1), we obtain
Therefore, the ratio of electric fields at the surface is
.
20. What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realize from your answer why ordinary capacitors are in the range of 
or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]
Ans.Capacitance of a parallel capacitor, V = 2 F
Distance between the two plates, d = 0.5 cm = 
Capacitance of a parallel plate capacitor is given by the relation,
Where,
= Permittivity of free space = 
Hence, the area of the plates is too large. To avoid this situation, the capacitance is taken in the range of
.
21.A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Fig. 2.36).
Show that the capacitance of a spherical capacitor is given by
where and 
are the radii of outer and inner spheres, respectively.
Ans.Radius of the outer shell =
Radius of the inner shell =
The inner surface of the outer shell has charge +Q.
The outer surface of the inner shell has induced charge 
Potential difference between the two shells is given by,
Where,
= Permittivity of free space
Capacitance of the given system is given by,
=
Hence, proved.
22. A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5
. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e., bending of field lines at the ends).
Ans.Length of a co-axial cylinder, l = 15 cm = 0.15 m
Radius of outer cylinder, 
= 1.5 cm = 0.015 m
Radius of inner cylinder, 
= 1.4 cm = 0.014 m
Charge on the inner cylinder, q = 3.5 
Capacitance of a co-axil cylinder of radii and
is given by the relation,
Where,
= Permittivity of free space = 
Potential difference of the inner cylinder is given by,
23. A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about
. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?
Ans.Potential rating of a parallel plate capacitor, V = 1 kV = 1000 V
Dielectric constant of a material, 
=3 Dielectric strength = 
For safety, the field intensity never exceeds 10% of the dielectric strength. Hence, electric field intensity, E = 10% of
= 
Capacitance of the parallel plate capacitor, C = 50 pF =
Distance between the plates is given by,
Capacitance is given by the relation,
Where,
A = Area of each plate
= Permittivity of free space =
Hence, the area of each plate is about 19
.