91Ó°ÊÓ

A Gaussian surface is a surface in which the electric field is the same at all points and the electric flux is always perpendicular to the surface.

A Gaussian surface is known as a closed surface in three-dimensional space, so the flux of the vector field is calculated. These vector fields can be either gravitational fields, electric fields or magnetic fields.

A spherical metal shell $r_1$has a charge $Q$(on its outer surface) and is surrounded by a concentric spherical metal shell $r_2$which has a charge $-Q$ (on its inner surface) is shown in the following figure.

In case of solid spheres, the net charge resides only on its surface. This means that the charge inside the solid sphere is zero. So, inside the solid sphere, the electric field is zero.

The electric potential due to a point charge$Q$ just outside a uniformly charged sphere is,

$V=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{Q}{r}$

At $r=r_1$(just outside from the inner shell), the electric potential can be expressed as,

$V_1=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{Q}{r_1}$

At $râ‰\yen r_2$, the electric potential can be expressed as,

$\begin{array}{rcl}V\text{'}& =& \frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{Q}{r}+\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{\left(-Q\right)}{r}\\ & =& 0\end{array}$

At $r=r_2$(just outside from the outer shell), the electric potential can be expressed as,

$V_2=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{Q}{r_2}$

Thus, the potential difference can be calculated as,

$\begin{array}{rcl}\mathrm{Δ}V& =& V_2-V_1\\ & =& \frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{Q}{r_2}-\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{Q}{r_1}\\ & =& \frac{Q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left(\frac{1}{r_2}-\frac{1}{r_1}\right)\\ & =& \frac{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{\left(\frac{1}{r_2}-\frac{1}{r_1}\right)}\mathrm{Δ}V\end{array}$

Thus, the capacitance of the spherical capacitor is,

role="math" localid="1662201260970" $\begin{array}{rcl}Q& =& C\mathrm{Δ}V\\ C& =& \frac{Q}{\mathrm{Δ}V}\\ & =& \frac{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{\left(\frac{1}{r_2}-\frac{1}{r_1}\right)}\left(\frac{\mathrm{Δ}V}{\mathrm{Δ}V}\right)\\ & =& \frac{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{\left(\frac{1}{r_2}-\frac{1}{r_1}\right)}\end{array}$

Hence, the capacitance of this spherical capacitor is $C=\frac{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{\left(\frac{1}{r_2}-\frac{1}{r_1}\right)}$.

Given that, the small gap distance between both plates of the spherical capacitor is,

$s=r_2-r_1$

The cross-sectional area of the plates of the spherical capacitor is,

localid="1662201620173" $A=4\mathrm{Ï€}{r}^2$

The capacitance of the spherical capacitor is,

$\begin{array}{rcl}C& =& \frac{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{\left(\frac{1}{r_2}-\frac{1}{r_1}\right)}\\ & =& \frac{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{\left(\frac{r_1-r_2}{r_1{r}_2}\right)}\\ & =& 4\mathrm{Ï€}{\mathrm{Î\mu }}_0\left(\frac{r_1{r}_2}{r_1-r_2}\right)\\ & =& 4\mathrm{Ï€}{\mathrm{Î\mu }}_0\left(\frac{r_1{r}_2}{s}\right)\end{array}$

When the radii of the spherical shells localid="1662201737051" $r_1$ and $r_2$ are large and nearly equal $\left(r_1≈r_2\right)$, to each other, then the capacitor of the spherical shell can be written as,

localid="1662202133117" $\begin{array}{rcl}C& =& 4\mathrm{Ï€}{\mathrm{Î\mu }}_0\left(\frac{r_1{r}_2}{s}\right)\\ & =& 4\mathrm{Ï€}{\mathrm{Î\mu }}_0\left(\frac{r^2}{s}\right)\\ & =& \frac{{\mathrm{Î\mu }}_{0}A}{s}\end{array}$

Hence, its proved.