 # Capacitance

## Introduction

Sometimes when you switch on the fan, it doesn’t move.

You try to rotate the fan by giving an external force or a torque via your hands to make it rotate then you call an electrician to get this issue resolved. The electrician tells us that the condenser isn’t functioning properly. Well, this condenser is nothing but the Capacitor.

• Capacitance

• Unit of capacitance

• SI unit of capacitance

• Unit of electrical capacitance

### Capacitance

The capacitance of a capacitor is defined as the ratio of the charge on the capacitor to the potential of the capacitor.

The electric field in the region between the conductors is directly proportional to the charge Q.

The potential difference B = Work done per unit positive charge in taking small test charge against the electric field.

Therefore, B is proportional to Q.

Hence the ratio Q /B  must be constant,

i.e.,

 Q/ B = constant = H

The constant H is called the Capacitance of the capacitor and it doesn’t depend upon Q and B.

The capacitance depends on the shape, size, and geometrical placing of the conductors and the medium between them.

### Potential of A Capacitor

The charge spread on the positive plate is called the charge on the capacitor.

The potential difference between these two plates is called the potential of a capacitor.

A capacitor is a combination of two metallic plates separated by an insulating medium where the magnitude of the positive charge spread on one plate (i.e. positive plate) equals the magnitude of the negative charge on the other plate (i.e. negative plate).

Let’s say the charge induced on one plate is +P and the potential difference is A+  and on the other, it is -N and a potential difference developed is A’

So, the net charge on the capacitor will be: Q = +P - N = 0.

The potential difference between these two plates: B = A - A’

The term charge doesn’t mean the total charge on a capacitor.

### Unit of capacitance

S.I unit of capacitance (H) is coulomb/volt which is written as farad.

If Q = 1 coulomb and

V = 1 volt,

 S.I. the base unit of H = s4⋅A2⋅m−2⋅kg−1Dimensional formula for H = [ M^-1 L^-2 T^4 I^2 ]

The capacitance of a conductor is said to be one farad when one coulomb of charge raises its potential via one volt.

This symbol F or farad is a large unit on normal scales and microfarad (mF) is used more frequently.

## Smaller Units of Capacitance

 1mF =  10 ^ - 6 farad1 nF = 1nF = 10 ^ - 9 farad1 mF = 1 m mF = 1 pf = 10 ^ -12 farad

The capacitance of a spherical conductor

An isolated charge conducting sphere has a capacitance which means a charged sphere has stored some energy as a result of being charged.

The potential of the inner sphere:

Ba = Q / 4 x π x μo x e - Q / 4 x π x μo x f

Here, Ba is the potential generated at the inner sphere A.

e =  Radius of the inner sphere.

μo =  Permittivity of free space (The proportionality constant that specifies the strength of the electric force between electric charge in a vacuum).

f =  Radius of the outer sphere B.

The outer sphere B is earthed so potential at Bb = 0.

Value of μo = 8.85 x 10 ^ -12 C ^2 N^ -1 m ^ - 2

1/  4 x π x μo = 9 x 10 ^ 9

The potential difference, B = Ba - Bb

=  Q x (1/ 4 x π x μo x e  - 1/ 4 x π x μo x f)

=  Q/ 4 x π x μo x (f - e) / e x f

Since B = Q / 4 x π x μo and Q = H x B

For an isolated spherical sphere:

H = 4 x π x μo/ [ 1/ e - 1/ f], and taking e as R and b -----> ∞

We get that:

 H = 4 x π x μo x R

The Capacitance of The Cylindrical Capacitor

A cylindrical capacitor consists of two co-axial cylinders of the same length l and radius R1 and R2.

The outer cylinder of radius R2 is earthed and the inner cylinder of radius R1 is given some charge.

The capacity of a cylindrical air conductor is given by

 H = 2 x π x μo x l / loge (R2 / R1)

The capacitance of the parallel plate capacitor

The electrical intensity (S) between two plates is given by

S = σ / A = 1 / μo Q / A x d

Since  B = S x d = 1/ μo x Q/ A x d

H = Q / B = Q / 1/ μo x Q/ A x d = μo x A/ d

.

Where H =capacitance

B= The potential difference between the two plates

Q =  The charge generated at plate 1 and plate 2.

σ = Surface charge density

A =  Area of each of the two plates separated by a distance d.

 H = μo x A/ d

### SAQs

Q1: Calculate the capacity of Earth taking it as a sphere of radius 6500km.

Solution: To find M

R = 6500 km

H =  4 x π x μo x R = 1/ 9 x 10 ^ 9 x 6.5 x 10 ^ 6

 H = 7.2 x 10 ^ - 4 F

Q2: Calculate the capacitance (H) of a parallel plate capacitor with the area of each plate 2 cm ^ 2 and separation of 4 mm.

Solution: Here, A = 2 cm ^ 2 = 10 ^ - 4  m ^ 2

d = 4 mm = 4 x 10 ^ - 3  m

μo = 8.85 x 10 ^ -12 C ^2 N^ -1 m ^ - 2

H = μo x A / d = 8.85 x 10 ^ -12 x  2 x 10 ^ - 4 / 4 x 10 ^ - 3

On solving we get:

 H = 4.425 x 10 ^ - 13 F =  4,425 x 10 ^ - 7 mF

### Summary

We observed that when the capacitor is given a charge through an external source or a battery, the charge accumulates on the two plates, and as soon as we connect this capacitor to the circuit the electric current ‘i’ starts flowing in the opposite direction of the electrons. The negative charge would towards the positive charge, and the charge on the plate would get neutralized.