# Spherical Capacitor

## What is a spherical Capacitor?

A capacitor consisting of two concentric spherical shells is called a spherical capacitor.

### Electric Field between spherical surfaces

Consider a spherical capacitor as shown in figure. Let,

• Concentric inner and outer spherical shells are ( A ) \ \text {and} \ ( B ) respectively.
• Radii of shells are ( a ) \ \text {and} \ ( b ) .
• Charge on inner shell is ( + q ) and outer shell is ( - q ) .
• The outer shell is earthed.

A Gaussian surface is drawn as shown in figure in dotted circle. Let, the radius of the spherical Gaussian surface is ( r ) . Then, according to Gauss law

\oint \vec {E} . d \vec {S} = \left ( \frac {q}{\epsilon_0} \right )

Because, ( \vec {E} ) and ( \vec {S} ) are along the same line, so ( \theta = 0 \degree ) . Therefore, \quad \oint \vec {E} . d \vec {S} = \oint E . d S \cos 0 \degree = \oint E . dS

Or, \quad \oint E . dS = \left ( \frac {q}{\epsilon_0} \right )

Since, ( E ) is uniform over the Gaussian surface due to symmetry. So, \quad E \oint d S = \left ( \frac {q}{\epsilon_0} \right )

Also, \quad \oint d S = \left ( 4 \pi r^2 \right ) ( Surface area of Gaussian surface ). Therefore, \quad E \times \left ( 4 \pi r^2 \right ) = \left ( \frac {q}{\epsilon_0} \right )

Or, \quad E = \left ( \frac {1}{4 \pi \epsilon_0} \right ) \left ( \frac {q}{r^2} \right ) ……….. (1)

Thus, intensity of electric field at a point of spherical capacitor is – (1) Proportional to the charge on the spherical surface and (2) inversely proportional to the square of the distance of point from the centre of spheres.

### Potential difference between spherical surfaces

From relation between electric field and potential difference, we get –

\vec E = - \left ( \frac {dV}{\vec {dr}} \right )

Or, \quad dV = - \vec {E} \ ( \vec {dr} )

By integration, we get –

V = - \int\limits_{b}^{a} \vec {E} . d \vec {r}

Since, ( \vec {E} ) and ( \vec {r} ) are along the same line. So, \quad \vec {E} . d \vec {r} = E. dr \cos 0 \degree = E dr

Therefore, \quad V = - \int\limits_{b}^{a} E. dr = - \int\limits_{b}^{a} \left ( \frac {1}{4 \pi \epsilon_0} \right ) \left ( \frac {q}{r^2} \right ) dr

= \left ( \frac {- q}{4 \pi \epsilon_0} \right ) \int\limits_{b}^{a} r^{(- 2)} dr

= \left ( \frac {- q}{4 \pi \epsilon_0} \right ) \left [ \frac {r^{(-1)}}{(-1)} \right ]_{b}^{a}

= \left ( \frac {q}{4 \pi \epsilon_0} \right ) \left [ \frac {1}{r} \right ]_{b}^{a}

= \left ( \frac {q}{4 \pi \epsilon_0} \right ) \left [ \frac {1}{a} - \frac {1}{b} \right ]

= \left [ \frac {q ( b - a )}{4 \pi \epsilon_0 ab} \right ]

= \left ( \frac {q}{4 \pi \epsilon_0} \right ) \left [ \frac {( b - a )}{ab} \right ] ……….. (2)

Thus, potential difference between spherical surfaces is – (1) Proportional to the charge on the spherical surface and (2) proportional to the difference of inverse of radii of the spheres.

### Capacitance of spherical Capacitor

By equation (2), the capacitance of spherical capacitor will be –

C = \left ( \frac {q}{V} \right )

= \left [ \frac {q}{\frac {q ( b - a )}{4 \pi \epsilon_0 ab}} \right ]

= \left [ \frac {4 \pi \epsilon_0 q ab}{q ( b - a )} \right ]

= 4 \pi \epsilon_0 \left [ \frac {ab}{( b - a )} \right ]

Therefore, capacitance of an spherical capacitor only depends upon the inner and outer radii of spheres.

## Cylindrical Capacitor

A capacitor consisting of two concentric cylinders is called a cylindrical capacitor.

Consider a cylindrical capacitor as shown in figure. Let, –

• Inner and outer cylindrical shells of capacitor are ( A ) \ \text {and} \ ( B ) .
• Length of cylinders are ( L ) and radii are ( a ) and ( b ) respectively.
• Inner cylinder surface charge is ( + q ) and outer cylinder surface charge is ( - q ) .
• The outer cylindrical surface is earthed as shown in figure.

### Electric Field between cylindrical surfaces

A Gaussian surface is drawn as shown in figure. According to Gauss law –

\oint \vec {E} . d \vec {S} = \left ( \frac {q}{\epsilon_0} \right )

So, \quad \oint E . d S \cos 0 \degree = \left ( \frac {q}{\epsilon_0} \right )

Or, \quad \oint E. dS = \left ( \frac {q}{\epsilon_0} \right )

Since, ( \vec {E} ) and ( \vec {S} ) are along the same line, so ( \theta = 0 \degree )

Also, ( E ) is uniform over the Gaussian surface. So, \quad E \oint d S = \left ( \frac {q}{\epsilon_0} \right )

But, \quad \oint d S = \left ( 2 \pi r L \right ) (Area of curved surface of Gaussian cylinder).

Therefore, \quad E \times \left ( 2 \pi r L \right ) = \left ( \frac {q}{\epsilon_0} \right )

Or, \quad E = \left ( \frac {q}{2 \pi \epsilon_0 r L} \right ) ………. (1)

Thus, intensity of electric field at a point of cylindrical capacitor is – (1) Proportional to the charge on the cylindrical surface and (2) inversely proportional to the distance of point from the centre of spheres and length of cylinder.

### Potential difference between cylindrical surfaces

The potential difference between two cylindrical shells of radii ( a ) and ( b ) with charge ( + q ) and ( - q ) respectively is given by –

V = - \int\limits_{b}^{a} E . d r = \left ( \frac {q}{2 \pi \epsilon_0 L} \right ) \int\limits_{a}^{b} \frac {dr}{r}

= \left ( \frac {q}{2 \pi \epsilon_0 L} \right ) \left [ \log_e r \right ]_{a}^{b}

= \left ( \frac {q}{2 \pi \epsilon_0 L} \right ) \left [ \log_e b - \log_e a \right ]

Therefore, \quad V = \left ( \frac {q}{2 \pi \epsilon_0 L} \right ) \log_e \left ( \frac {b}{a} \right ) ……. (2)

### Capacitance of cylindrical Capacitor

Using equation (2), capacitance of a cylindrical capacitor will be –

C = \left ( \frac {q}{V} \right )

= \left [ \frac {q }{\frac {q}{2 \pi \epsilon_0 L} \log_e \left ( \frac {b}{a} \right )} \right ]

= \left [ \frac {2 \pi \epsilon_0 L}{\log_e ( b / a )} \right ]

= \left [ \frac {2 \pi \epsilon_0 L}{2.303 \log_{10} ( b / a )} \right ] ……… (3)

Therefore, capacitance of a cylindrical capacitor only depends upon the length and inner and outer radii of cylindrical shells.