Name: Dielectrics in a Capacitor
Uploaded: 2019-07-19
Duration: 6 min 3 s
Description: Let us use our knowledge about dielectrics and capacitors to find out how the use of dielectrics can increase the capacitance of a system. Learn through this...

Question

A capacitor consists of two air spaced concentric cylinders. The outer radius $b$ is fixed, and the inner is of radius $a$ . If breakdown of air occurs at field strength than $E b$ show that the inner cylinder should have

(i) radius $a = \frac{b}{e}$ if the potential of the inner cylinder is to be maximum.

(ii) radius $a = \frac{b}{\sqrt{e}}$ if the energy per unit length of the system is to be maximum.

Accepted Answer

Let the inner surface of the outer shell has charge $+ Q$ ,

The outer surface of the inner shell has induced charge $- Q .$

Given:- Radius of the outer shell $= b$ , Radius of the inner shell $= a$ ,

Therefore, Potential difference between the two shells is given by, $\frac{Q}{4 {πε}_{\circ} a} - \frac{Q}{4 {πε}_{\circ} b}$

Where, $ε_{\circ} =$ Permittivity of free space

$V = \frac{1}{4 {πε}_{\circ}} [\frac{1}{a} - \frac{1}{b}]$

Case :-1

Radius $a = \frac{b}{e}$ if the potential of the inner cylinder is to be maximum.

$∴ V = \frac{1}{4 {πε}_{\circ}} [\frac{1}{\frac{b}{e}} - \frac{1}{b}]$

$∴ V = \frac{1}{4 {πε}_{\circ}} [\frac{e}{b} - \frac{1}{b}]$

$∴ V = \frac{1}{4 {πε}_{\circ}} [\frac{e - 1}{b}]$

Case :-2

radius $a = \frac{b}{\sqrt{e}}$ if the energy per unit length of the system is to be maximum.

$∴ V = \frac{1}{4 {πε}_{\circ}} [\frac{1}{\frac{b}{\sqrt{e}}} - \frac{1}{b}]$

$∴ V = \frac{1}{4 {πε}_{\circ}} [\frac{\sqrt{e}}{b} - \frac{1}{b}]$

$∴ V = \frac{1}{4 {πε}_{\circ}} [\frac{\sqrt{e} - 1}{b}]$

Important Questions on Capacitance

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