In This Post we are providing Chapter-2 ELECTROSTATIC POTENTIAL AND CAPACITANCE NCERT MOST IMPORTANT QUESTIONS for Class 12 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter
NCERT MOST IMPORTANT QUESTIONS ON ELECTROSTATIC POTENTIAL AND CAPACITANCE
1. Explain what would happen if in the capacitor , 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 capacitance,
Supply voltage, V = 100 V
New charge
sPotential 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,
2. 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 el
ectrostatic energy stored in the capacitor is 
3. 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
.
4. 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.
5. 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
.
6. 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
.
7. 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
.
8.A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports
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.
9. 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,
10. 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
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