Please scroll down to get to the study materials.

- Note
- Things to remember
- Videos

A capacitor is an electric device used to store electric charge. It is consists of two conducting surface separated, at a small distance, called the plates of the capacitor. One plate stores positive charge and another holds the negative charge. The ability of a capacitor to store electric charge is called its capacitance.

The potential difference between two plates of a capacitor is directly proportional to the charge stored on it.

$$\text {i.e.} q \propto V$$

$$\text {or,} q = V$$

where C is a proportionality constant called capacitance of the capacitor. Its value depends upon the geometry of the capacitor and medium present between the plates.

$$\therefore C = \frac{q}{V}$$

unit of \(C = \frac {\text{columb(c)}}{\text{volt(V)}} = Farad (F)\)

Hence, the capacitance of a capacitor is numerically equal with the charge raise the potential difference between the plates of the capacitor by 1 volt.

**One Farad Capacitance**

We have, \(q=CV \)

$$C= \frac{q}{V}$$

$$\text{for,} q= 1C, V=1 volt$$

$$C = \frac{1C}{1V} = 1 Farad$$

Hence, the capacitance of a capacitor is said to be 1 Farad

Hence, the capacitance of a capacitor is said to be 1 Farad if 1 Columb charge is required to increase the potential difference between its plates by 1 volt.

Consider an isolated spherical conductor in air or vacuum. Let its radius be R and charge on its surface Q.

The potential at any point on its surface is given by

$$V= \frac{Q}{4\pi \epsilon _o R}$$

$$\text{or} \frac {Q}{V}4\pi \epsilon _o R $$

where \(\epsilon _o\) is the permittivity of air or free space as the sphere is placed in air.

As, \( \frac {Q}{V} = C ,\) the capacitance of an isolated charged sphere. The second conductor is a sphere of an infinite radius where the potential is zero.

In CGS-system, \(4\pi \epsilon _o = 1\) and

$$ C = R $$

So, in CGS_system, the capacitance of the isolated charged sphere is numerically equal to its radius.

**Spherical Capacitor**

A spherical capacitor consists of two concentric hollow metal spheres of different radii as shown in the figure. Let +Q be the charge on the surface of the inner sphere A and the outer space B be earthed. Due to the electrostatic induction, an equal but opposite charge -Q is induced on the inner surface of B is earthed, the +Q charge of the outer sphere flows to the earth.

Let a and b be the radius of the sphere A and the sphere B respectively, and there be a vacuum or air in the space between the two spheres.

The potential of inner sphere,

$$V_a = \text{(potential due to + Q charge)} + \text{(Potential due to - Q charge)}$$

$$= \frac{Q}{4\pi \epsilon _o a} + \frac{(-Q)}{4\pi \epsilon _o b}$$

$$= \frac{Q}{4\pi \epsilon _o a} - \frac{(Q)}{4\pi \epsilon _o b}$$

$$= \frac{Q}{4\pi \epsilon _o} \left [\frac{1}{a} - \frac{1}{b}\right] $$

As the outer space is earthed, the potential due to the outer sphere V_{b} = 0. So, the potential difference between the inner and outer sphere is

$$V = V_a - V_b$$

$$ =\left( \frac{Q}{4\pi \epsilon _o a} - \frac{(Q)}{4\pi \epsilon _o b}\right ) - 0$$

$$= \frac{Q}{4\pi \epsilon _o} \left [\frac{1}{a} - \frac{1}{b}\right] $$

$$= \frac{Q}{4\pi \epsilon _o} \left (\frac{1}{a} \frac{b - a}{ab}\right)$$

$$\text{or,} \frac{Q}{V} = 4\pi \epsilon _o \left (\frac{ab}{b-a}\right) $$

So, the capacitance, \(C =4\pi \epsilon _o \left (\frac{ab}{b-a}\right) \)

If we have an isolated charged sphere of radius a, its capacitance is\(C =4\pi \epsilon _o a\). Since \(\frac {ab}{b-a} > a\), it follows that C>C'. Therefore, the arrangement of the two spherical shells leads to increase the capacitance of a spherical conductor.

A parallel plate capacitor consists of two conducting parallel plates separated at a small distance and a dielectric material is present between them.

Consider a parallel plate capacitor having plates P and Q with the plate with plate area 'A' separated at a distance 'd'. If the surface charge density is \('\sigma'\) then the electric field intensity at a point between the plates is given by

$$E = \frac{\sigma}{\epsilon _o}\dots (i)$$

If charge 'q' is contained on the plate, then by the definition of charge density is \('\sigma'\)

$$\sigma = \frac {q}{A}\dots (ii)$$

using equation (i) and (ii), we get,

$$E = \frac{q}{\epsilon _o A}\dots (iii)$$

If the potential difference between the plates of the capacitor is 'V', then the electric field intensity (E) in terms of potential difference can be written as

$$E = \frac{V}{d}\dots (iv)$$

using equation (iii) and (iv), we get,

$$\frac{V}{d} = \frac{q}{\epsilon _o A}$$

$$\text{or,} q = \left( \frac {\epsilon _o A}{d}\right) V \dots (v)$$

But we know that

$$q = C.V \dots(vi)$$

using equation (iv) and (v), we get,

$$C = \frac{\epsilon _o A}{d}$$

This gives the capacitance of the parallel capacitor.

If a medium having permittivity \(\epsilon\) is placed between the plates, then the capacitance of the capacitor becomes

$$C_m = \frac{\epsilon .A}{d}$$

Hence, by increasing the area of the plates(A) permittivity it the medium between the plates\(\epsilon\) and reducing the distance between the plates of the capacitor, its capacitor can be increased.

A capacitor is an electric device used to store electric charge.

The capacitance of a capacitor is numerically equal with the charge raise the potential difference between the plates of the capacitor by 1 volt.

The capacitance of a capacitor is said to be 1 Farad if 1 Columb charge is required to increase the potential difference between its plates by 1 volt.

A parallel plate capacitor consists of two conducting parallel plates separated at a small distance and a dielectric material is present between them.

Increasing the area of the plates(A) permittivity it the medium between the plates and reducing the distance between the plates of the capacitor, its capacitor can be increased.

.-
## You scored /0

## ASK ANY QUESTION ON Capacitance of a Capacitor

No discussion on this note yet. Be first to comment on this note