Question

In cylindrical capacitor, electrical field is maximum at

• A. inner radius
• B. outer radius
• C. between inner and outer radius
• D. none of these

Answer 1

The correct option is A inner radius

If $V$ is the potential difference between inner and outer shells of radii $a,b$ respectively. Hence, capacitance of cylindrical capacitor will be

$C=q/V$

Let us draw a gaussian cylindrical surface around inner shell, of radius $r$, then by Gauss's theorem, total electric flux through gaussian surface,

$\varphi =q/{\epsilon }_0$ .........eq1 ($q=$ charge on inner shell)

Now, if magnitude of electric field at gaussian surface is $E$ ,then

$\varphi =E\times 2\pi rL$ ............eq2

equating eq1 and eq2 ,

$E\times 2\pi rL=q/{\epsilon }_0$

or $E=\frac{q}{2\pi {\epsilon }_{0}rL}$ ................eq3

From above equation,

$E\propto 1/r$, and the minimum value of $r$ is $a$.

Hence, electric field is maximum at inner radius.