Capacitance of a Capacitor

Capacitor

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.

Capacitance of different capacitor(C)

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.

Isolated Sphere

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.

Capacitance of a Parallel Capacitor

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.