Chapter 2: Electrostatic Potential and Capacitance

Given,

q₁ = 5 x 10^-8 C

q₂ = -3 x 10^-8 C

The two charges are at a distance, d = 16 cm = 0.16 m from each other.

Let us consider a point “P” over the line joining charges q₁ and q₂.

Let the distance of the considered point P from q₁ be ‘r’

Let us consider point P to have zero electric potential (V).

The electric potential at point P is the summation of potentials due to charges q₁ and q₂.

Therefore, 

Here,

Putting  V = 0, in equation (1), we get,

0 =

5(0.16 – r)  = 3r

0.8 = 8r

r = 0.1 m = 10 cm.

Therefore, at a distance of 10 cm from the positive charge, the potential is zero between the two charges.

Let us assume a point P at a distance ‘s’ from the negative charge is outside the system, having a potential zero.

So, for the above condition, the potential is given by

Here,

For V = 0, equation (2) can be written as :

0 =

5(s – 0.16) = 3s

0.8  = 2s

S = 0.4 m = 40 cm.

Therefore, the potential is zero at a distance of 40 cm from the positive charge outside the system of charges.

The figure shows a regular hexagon containing charges q at each of its vertices.

Here,

q =  5 µC = 5 × 10^-6 C.

Length of each side of the hexagon, AB =BC = CD = DE = EF = FA = 10 cm.

The distance of the vertices from the centre O, d = 10 cm.

The electric potential at point O,

The electric potential, V = 

Here,

V = 

(1) An equipotential surface is defined as the surface over which the total potential is zero. In the given question, the plane is normal to line AB. The plane is located at the mid-point of line AB as the magnitude of the charges is the same.

(2) At every point on this surface, the direction of the electric field is normal to the plane in the direction of AB.

(1) Given,

Radius of spherical conductor, r = 12cm = 0.12m

The charge is distributed uniformly over the surface, q = 1.6 x 10^-7 C.

The electric field inside a spherical conductor is zero.

(2) Electric field E, just outside the conductor, is given by the relation,

E = 

Here,

ε₀ = permittivity of free space and

Therefore,

E =

Therefore, just outside the sphere, the electric field is 10⁵ NC^-1.

(3) From the centre of the sphere, the electric field at a point 18m = E₁.

From the centre of the sphere, the distance of point d = 18 cm = 0.18m.

E₁ =

So, from the centre of the sphere, the electric field at a point 18 cm away is 4.4 x 10⁴ NC^-1.

Given,

Capacitance, C = 8pF.

In the first case, the parallel plates are at a distance ‘d’ and are filled with air.

Air has a dielectric constant, k = 1

Capacitance, C =

Here,

A = area of each plate

Now, if the distance between the parallel plates is reduced to half, then d₁ = d/2

Given the dielectric constant of the substance, k₁ = 6

Hence, the capacitance of the capacitor,

C₁ =

Taking ratios of equations (1) and (2), we get,

C₁ = 2 x 6 C = 12 C = 12 x  8 pF = 96pF.

Hence, the capacitance between the plates is 96pF.

(1)  Given,

The capacitance of the three capacitors, C = 9 pF

Equivalent capacitance (c_eq) is the capacitance of the combination of the capacitors given by

Therefore, the total capacitance = 3pF.

(2) Given supply voltage, V = 100V

The potential difference (V₁) across the capacitors will be equal to one-third of the supply voltage.

Therefore, V₁ = 

Hence, the potential difference across each capacitor is 40V.

(1) Given, C₁ = 2pF, C₂ = 3pF and C₃ = 4pF.

Equivalent capacitance for the parallel combination is given by C_eq.

Therefore, C_eq = C₁ + C₂ + C₃ = 2 + 3 + 4 = 9pF

Hence, the total capacitance of the combination is 9pF.

(2) Supply voltage, V = 100V

The three capacitors have the same voltage, V = 100v

q = VC

where,

q = charge

C = capacitance of the capacitor

V = potential difference

for capacitance, c = 2pF

q = 100 x 2 = 200pC = 2 x 10^-10C

for capacitance, c = 3pF

q = 100 x 3 = 300pC = 3 x 10^-10C

for capacitance, c = 4pF

q = 100 x 4 = 400pC = 4 x 10^-10 C

Solution:  Given,

The area of the plate of the capacitor, A = 6 x 10^-3 m²

Distances between the plates, d = 3mm = 3 x 10^-3 m

The voltage supplied, V = 100V

Capacitance of a parallel plate capacitor is given by, C = 

Here,

ε_o = permittivity of free space = 8.854 × 10^-12 N^-1 m ^-2 C^-2

C =

Therefore, each plate of the capacitor has a charge of

q = VC = 100 × 17.71 x 10^-12 C = 1.771 x 10^-9 C

Solution: Given,

Capacitance of the capacitor, C = 12pF = 12 x 10^-12 F

Potential difference, V = 50 V

Electrostatic energy stored in the capacitor is given by the relation,

E =

Therefore, the electrostatic energy stored in the capacitor is 1.5 x 10^-8 J.

Solution: Given,

Capacitance, C = 600pF

The potential difference, V = 200v

Electrostatic energy stored in the capacitor is given by :

E₁ =

According to the question, the source is disconnected from the 600pF and connected to another capacitor of 600pF, then equivalent capacitance (C_eq) of the combination is given by,

=

C_eq = 300pF

New electrostatic energy can be calculated by:

E₂ =

Loss in electrostatic energy,

E = E₁ – E₂

E = 1.2 x 10^-5 – 0.6 x 10^-5 J = 0.6 x 10^-5 J = 6 x 10^-6 J

Therefore, the electrostatic energy lost in the process is 6 x 10^-6 J.