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 5 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
5 Marks Questions
1. (a) Define dielectric constant in terms of the capacitance of a capacitor? On what factor does the capacitance of a parallel plate capacitor with dielectric depend?
(b) Find the ratio of the potential differences that must be applied across the
(1) parallel
(2) Series combination of two identical capacitors so that the energy stored in the two cases becomes the same.
Ans. (a) Dielectric constant is defined as the ratio of capacitance of a capacitor when the dielectric is filled in between the plates to the capacitance of a capacitor when there is vaccuum in between the plates
In K = 
Capacitance of a parallel plate capacitor with dielectric depends on the following factors 
(1) Area of the plates
(2) Distance between the plates
(3) Dielectric constant of the dielectric between the plates
(b) Let the capacitance of each capacitor be C
Cp = C + C = 2C (in parallel)
CS = 
(in series)
Let 
be the values of potential difference
This 
But 
(given)
Vp : Vs = 1 : 2
2. (a) An air filled capacitor is given a charge of 2
C raising its potential to 200 V. If on inserting a dielectric medium, its potential falls to 50 V, what is the dielectric constant of the medium?
(b) A conducting stab of thickness ‘t’ is introduced without touching between the
plates of a parallel plate capacitor separated by a distance d (t<d). Derive an
expression for the capacitance of a capacitor?
Ans. (a) 
(Where V = 200 V for air filled capacitor 
after insertion of a dielectric)
(b) For a parallel plate capacitor when air/vaccuum is in between the plates 
Since electric field inside a conducting stab is zero, hence electric field exist only between the space (d-t) 
Where 
is the electric field exist between the plates?
And 
Where A is the area of each plates
Hence capacitor of a parallel plate capacitor
3. Figure (a) and (b) shows the field lines of a single positive and negative changes respectively
(a) Give the signs of the potential difference: 
(b) Give the sign of the work done by the field in moving a small positive change from Q to P.
(c) Give the sign of the work done by the field in moving a small negative change from B to A.
Ans. (a) We know 
Because
is less negative then
.
(b) In moving a positive change form Q to P work has to be done against the electric field so it is negative.
(c) In moving a negative change form B to A work is done along the same direction of the field so it is positive.
4. With the help of a labelled diagram, explain the principle, construction and working of a vandegraff generator. Mention its applications?
Ans. Vandegraff generator is a device which is capable of producing a high potential of the order of million volts.
Principle (1) The charge always resides on the outer surface of hollow conductor.
(2) The electric discharge in air or gas takes place readily at the pointed ends of the conductors.
Construction: – It consist of a large hollow metallic sphere S mounted on two insulating columns A and B and an endless belt made up of rubber which is running over two pulleys
with the help of an electric motor. 
are two sharp metallic brushes. The lower brush 
is given a positive potential by high tension battery and is called a spray brush while the upper brush 
connected to the inner part of the Sphere S.
Working: – When brush is given a high positive potential then it produces ions, due to the action of sharp points. Thus the positive ions so produced get sprayed on the belt due to repulsion between positive ions and the positive charge on brush. Then it is carried upward by the moving belt. The pointed end ofjust touches the belt collects the positive change and make it move to the outer surface of the sphere S. This process continues and the potential of the shell rises to several million volts.
Applications – Particles like proton, Deutrons, 
— particles etc are accelerated to high speeds and energies.
5. Two charges 
and 
are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
Ans.There are two charges,
Distance between the two charges, d = 16 cm = 0.16 m
Consider a point P on the line joining the two charges, as shown in the given figure.
r = Distance of point P from charge
Let the electric potential (V) at point P be zero.
Potential at point P is the sum of potentials caused by charges q1 and q2 respectively.
Where,
= Permittivity of free space
For V = 0, equation (i) reduces to
Therefore, the potential is zero at a distance of 10 cm from the positive charge between the charges.
Suppose point P is outside the system of two charges at a distance s from the negative charge, where potential is zero, as shown in the following figure.
For this arrangement, potential is given by,
For V = 0, equation (ii) reduces to
Therefore, the potential is zero at a distance of 40 cm from the positive charge outside the system of charges.
6. A parallel plate capacitor with air between the plates has a capacitance of 8 pF (
What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?
Ans. Capacitance between the parallel plates of the capacitor, C = 8 pF
Initially, distance between the parallel plates was d and it was filled with air. Dielectric constant of air, k = 1
Capacitance, C, is given by the formula,
Where,
A = Area of each plate
= Permittivity of free space
If distance between the plates is reduced to half, then new distance, 
Dielectric constant of the substance filled in between the plates, 
= 6 Hence, capacitance of the capacitor becomes
…….(ii)
Taking ratios of equations (i) and (ii), we obtain
Therefore, the capacitance between the plates is 96 pF.
7. A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of 
from a point P (0, 0, 3 cm) to a point Q (0, 4 cm, 0), via a point R
(0, 6 cm, 9 cm).
Ans. Charge located at the origin, q = 8 mC
Magnitude of a small charge, which is taken from a point P to point R to point Q, 
All the points are represented in the given figure.
Point P is at a distance, 
, from the origin along z-axis. Point Q is at a distance, 
, from the origin along y-axis.
Potential at point P, 
Potential at point Q, 
Work done (W) by the electrostatic force is independent of the path.
Therefore, work done during the process is 1.27 J.
8. A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.
Ans.Length of the side of a cube = b
Charge at each of its vertices = q
A cube of side b is shown in the following figure.
d = Diagonal of one of the six faces of the cube
l = Length of the diagonal of the cube
Is the difference between the centre of the cube and one of the eight vertices
The electric potential (V) at the centre of the cube is due to the presence of eight charges at the vertices.
Therefore, the potential at the centre of the cube is .
The electric field at the centre of the cube, due to the eight charges, gets cancelled. This is because the charges are distributed symmetrically with respect to the centre of the cube. Hence, the electric field is zero at the centre.
9. Two tiny spheres carrying charges 1.5 
are located 30 cm apart. Find the potential and electric field:
1. at the mid-point of the line joining the two charges, and
2. at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.
Ans.
Two charges placed at points A and B are represented in the given figure. O is the mid- point of the line joining the two charges.
Magnitude of charge located at A, q1 = 1.
Magnitude of charge located at B, q2 = 2.
Distance between the two charges, d = 30 cm = 0.3 m
1. Let
are the electric potential and electric field respectively at O.
V1 = Potential due to charge at A + Potential due to charge at B
Where,
= Permittivity of free space
= Electric field due to 
Therefore, the potential at mid-point is 
and the electric field at mid-point is
. The field is directed from the larger charge to the smaller charge.
1. Consider a point Z such that normal distance OZ = 10 cm = 0.1 m, as shown in the following figure.
are the electric potential and electric field respectively at Z. It can be observed from the figure that distance,
BZ=AZ=
= Electric potential due to A + Electric Potential due to B
Electric field due to q at Z,
Electric field due to q2 at Z,
The resultant field intensity at Z,
Where, 
is the angle, 
From the figure, we obtain
Therefore, the potential at a point 10 cm (perpendicular to the mid-point) is 
and electric field is
.
10. Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by
Where 
is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is 
1. Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another. [Hint: For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]
Ans. Electric field on one side of a charged body is
and electric field on the other side of the same body is
. If infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by,
Where,
= Unit vector normal to the surface at a point σ = Surface charge density at that point Electric field due to the other surface of the charged body,
Electric field at any point due to the two surfaces,
Since inside a closed conductor, 
= 0,
Therefore, the electric field just outside the conductor is.
1. When a charged particle is moved from one point to the other on a closed loop, the work done by the electrostatic field is zero. Hence, the tangential component of electrostatic field is continuous from one side of a charged surface to the other.
11. A long charged cylinder of linear charged density 
is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?
Ans. Charge density of the long charged cylinder of length L and radius r is λ.
Another cylinder of same length surrounds the pervious cylinder. The radius of this cylinder is R.
Let E be the electric field produced in the space between the two cylinders.
Electric flux through the Gaussian surface is given by Gauss’s theorem as,
Where, d = Distance of a point from the common axis of the cylinders Let q be the total charge on the cylinder.
It can be written as
Where,
q = Charge on the inner sphere of the outer cylinder
= Permittivity of free space
Therefore, the electric field in the space between the two cylinders is .
12.In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 
.
1. Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton.
2. What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?
3. What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 separation?
Ans.The distance between electron-proton of a hydrogen atom, 
Charge on an electron, 
Charge on a proton, 
1. Potential at infinity is zero.
Potential energy of the system, p-e = Potential energy at infinity -Potential energy at distance, d
Where,
is the permittivity of free space
Potential enerygy
Since 
Potential energy
Therefore, the potential energy of the system is 
2. Kinetic energy is half of the magnitude of potential energy.
Kinetic energy
Total energy = 
= 13.6 eV
Therefore, the minimum work required to free the electron is 13.6 eV.
3. When zero of potential energy is taken, 
Potential energy of the system = Potential energy at 
– Potential energy at d
13. Two charges 
are located at points 
respectively.
1. What is the electrostatic potential at the points?
2. Obtain the dependence of potential on the distance r of a point from the origin when r/a >> 1.
3. How much work is done in moving a small test charge from the point (5, 0, 0) to 
along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?
Ans. 
Zero at both the points
Charge 
and charge + q is located at (0, 0, a). Hence, they form a dipole. Point (0, 0, z) is on the axis of this dipole and point (x, y, 0) is normal to the axis of the dipole. Hence, electrostatic potential at point (x, y, 0) is zero. Electrostatic potential at point (0, 0, z) is given by,
Where,
= Permittivity of free space
p = Dipole moment of the system of two charges = 2qa
2. Distance r is much greater than half of the distance between the two charges. Hence, the potential (V) at a distance r is inversely proportional to square of the distance
i.e., 
3. Zero
The answer does not change if the path of the test is not along the x-axis.
A test charge is moved from point (5, 0, 0) to point (-7, 0, 0) along the x-axis. Electrostatic potential 
at point (5, 0, 0) is given by,
Electrostatic potential, , at point (- 7, 0, 0) is given by,
Hence, no work is done in moving a small test charge from point (5, 0, 0) to point (−7, 0, 0) along the x-axis.
The answer does not change because work done by the electrostatic field in moving a test charge between the two points is independent of the path connecting the two points.
14. Figure 2.34 shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on r for r/a >> 1, and contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge).
Ans. Four charges of same magnitude are placed at points X, Y, Y, and Z respectively, as shown in the following figure.
A point is located at P, which is r distance away from point Y. The system of charges forms an electric quadrupole.
It can be considered that the system of the electric quadrupole has three charges.
Charge +q placed at point X
Charge 
placed at point Y
Charge +q placed at point Z
XY = YZ = a
YP = r
PX = r + a
PZ = 
Electrostatic potential caused by the system of three charges at point P is given by,
Since
,
is taken as negligible.
It can be inferred that potential, 
However, it is known that for a dipole, 
And, for a monopole,
15. An electrical technician requires a capacitance of 
in a circuit across a potential difference of 1 kV. A large number of 1 2µF capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors.
Ans. Total required capacitance, C = 
Potential difference, V = 1 kV = 1000 V Capacitance of each capacitor, 
= 
Each capacitor can withstand a potential difference, 
= 400 V
Suppose a number of capacitors are connected in series and these series circuits are connected in parallel (row) to each other. The potential difference across each row must be 1000 V and potential difference across each capacitor must be 400 V. Hence, number of capacitors in each row is given as
Hence, there are three capacitors in each row. Capacitance of each row 
+…….n terms
However, capacitance of the ciruit is given as 2
.
N=6
Let there are n rows, each having three capacitors, which are connected in parallel. Hence, equivalent capacitance of the circuit is given as
Hence, 6 rows of three capacitors are present in the circuit. A minimum of 
i.e., 18 capacitors are required for the given arrangement.
16.Obtain the equivalent capacitance of the network in Fig. 2.35. For a 300 V supply, determine the charge and voltage across each capacitor.
Ans.Capacitance of capacitor is 100 pF.
Capacitance of capacitor 
is 200 pF.
Capacitance of capacitor 
is 200 pF.
Capacitance of capacitor 
is 100 pF.
Supply potential, V = 300 V
Capacitors and are connected in series. Let their equivalent capacitance be 
=100pF
Capacitors 
are in parallel. Let their equivalent capacitance be 
=100+100=200pF
are connected in series. Let their equivalent capacitance be C.
Hence, the equivalent capacitance of the circuit is 
Potential difference across 
Potential difference across = 
Charge on is given by,
= CV
voltage across is given by,
=300-200=100V
Hence, potential difference, , across is 100 V. Charge on is given by,
and having same capacitances have a potential difference of 100 V together. Since
and are in series, the potential difference across and is given by,
= 
= 50 V
Therefore, charge on is given by,
And charge on is given by,
pF with
Hence, the equivalent capacitance of the given circuit is
17. The plates of a parallel plate capacitor have an area of 90
each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply.
1. How much electrostatic energy is stored by the capacitor?
2. View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u. Hence arrive at a relation between u and the magnitude of electric field E between the plates.
Ans. Area of the plates of a parallel plate capacitor, A = 90 = 
Distance between the plates, d = 2.5 mm = 
Potential difference across the plates, V = 400 V
1. Capacitance of the capacitor is given by the relation,
Electrostatic energy stored in the capacitor is given by the relation, 
Where,
= Permittivity of free space = 
Hence, the electrostatic energy stored by the capacitor is 
2. Volume of the given capacitor,
Energy stored in the capacitor per unit volume is given by,
Again, u=
Where,
= Electric intensity = E
18. A 4 
capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2
capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?
Ans. Capacitance of a charged capacitor, 
Supply voltage, = 200 V Electrostatic energy stored in 
is given by,
F
Capacitance of an uncharged capacitor,
When is connected to the circuit, the potential acquired by it is 
.
According to the conservation of charge, initial charge on capacitor is equal to the final charge on capacitors, and .
Electrostatic energy for the combination of two capacitors is given by,
Hence, amount of electrostatic energy lost by capacitor
19. Show that the force on each plate of a parallel plate capacitor has a magnitude equal to 
QE, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor
.
Ans. Let F be the force applied to separate the plates of a parallel plate capacitor by a distance of
. Hence, work done by the force to do so 
As a result, the potential energy of the capacitor increases by an amount given as
.
Where,
u = Energy density
A = Area of each plate
d = Distance between the plates
V = Potential difference across the plates
The work done will be equal to the increase in the potential energy i.e.,
Electric intensity is given by,
However, capacitance, 
Charge on the capacitor is given by,
Q = CV
The physical origin of the factor, , in the force formula lies in the fact that just outside the conductor, field is E and inside it is zero. Hence, it is the average value, 
, of the field that contributes to the force.
20. A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 µC. The space between the concentric spheres is filled with a liquid of dielectric constant 32.
1. Determine the capacitance of the capacitor.
2. What is the potential of the inner sphere?
3. Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.
Ans. Radius of the inner sphere, 
= 12 cm = 0.12 m Radius of the outer sphere, 
= 13 cm = 0.13 m Charge on the inner sphere,
Dielectric constant of a liquid, 
(a) Capacitance of the capacitor is given by the relatioon,
Where,
= Permittivity of free space = 
Hence, the capacitance of the capacitor is approximately.
1. Potential of the inner sphere is given by,
Hence, the potential of the inner sphere is 
.
2. Radius of an isolated sphere, r = 
3. Capacitance of the sphere is given by the relation,
The capacitance of the isolated sphere is less in comparison to the concentric spheres. This is because the outer sphere of the concentric spheres is earthed. Hence, the potential difference is less and the capacitance is more than the isolated sphere.
21. Answer carefully:
1. Two large conducting spheres carrying charges
and 
are brought close to each other. Is the magnitude of electrostatic force between them exactly given by 
, where r is the distance between their centres?
2. If Columb’s law involved
dependence (instead of 
), would Gauss’s law be still true?
3. A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?
4. What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical?
1. We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?
2. What meaning would you give to the capacitance of a single conductor?
3. Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (= 6).
Ans. 1. The force between two conducting spheres is not exactly given by the expression
, because there is a non-uniform charge distribution on the spheres.
2. Gauss’s law will not be true, if Coulomb’s law involved 
dependence, instead of 
, on r.
3. Yes, If a small test charge is released at rest at a point in an electrostatic field configuration, then it will travel along the field lines passing through the point, only if the field lines are straight. This is because the field lines give the direction of acceleration and not of velocity.
1. Whenever the electron completes an orbit, either circular or elliptical, the work done by the field of a nucleus is zero.
2. No Electric field is discontinuous across the surface of a charged conductor. However, electric potential is continuous.
1. The capacitance of a single conductor is considered as a parallel plate capacitor with one of its two plates at infinity.
2. Water has an unsymmetrical space as compared to mica. Since it has a permanent dipole moment, it has a greater dielectric constant than mica.
22. Answer the following:
1. The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100
. Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside!)
2. A man fixes outside his house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area
. Will he get an electric shock if he touches the metal sheet next morning?
3. The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged?
4. What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning? (Hint: The earth has an electric field of about
at its surface in the downward direction, corresponding to a surface charge density =
. Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about + 1800 C is pumped every second into the earth as a
whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.)
Ans. We do not get an electric shock as we step out of our house because the original equipotential surfaces of open air changes, keeping our body and the ground at the same potential.
1. Yes, the man will get an electric shock if he touches the metal slab next morning. The steady discharging current in the atmosphere charges up the aluminum sheet. As a result, its voltage rises gradually. The raise in the voltage depends on the capacitance of the capacitor formed by the aluminium slab and the ground.
2. The occurrence of thunderstorms and lightning charges the atmosphere continuously. Hence, even with the presence of discharging current of 1800 A, the atmosphere is not discharged completely. The two opposing currents are in equilibrium and the atmosphere remains electrically neutral.
3. During lightning and thunderstorm, light energy, heat energy, and sound energy are dissipated in the atmosphere.
