The earth is assumed to be a charged conducting sphere having volume ' $\mathrm{V}^{\prime}$ and surface…
The earth is assumed to be a charged conducting sphere having volume ' $\mathrm{V}^{\prime}$ and surface area 'A'. The capacitance of the earth in free space is $\left(\mathrm{E}_{0}=\right.$ permittivity of free space $)$
$12 \pi \in_{0} \frac{\mathrm{V}}{\mathrm{A}}$
$4 \pi \in_{0} \frac{\mathrm{V}}{\mathrm{A}}$
$2 \pi \in_{0} \frac{\mathrm{V}}{\mathrm{A}}$
$8 \pi \in_{0} \frac{\mathrm{V}}{\mathrm{A}}$
Solution
If we assume shape of earth be solid spherical.
Then, the volume of earth $=4 / 3 \pi \mathrm{R}^{3}$
where $R$ is radius of earth.
Given, volume of earth $=\mathrm{V}=4 / 3 \pi \mathrm{R}^{3}$
surface area of earth $=\mathrm{A}=4 \pi \mathrm{R}^{2}$
So, $V / A=\left(4 / 3 R^{3}\right) / 4 \pi R^{2}$
or $\mathrm{V} / \mathrm{A}=\mathrm{R} / 3$
or, $\mathrm{R}=3 \mathrm{~V} / \mathrm{A}$
Now, capacitance of earth $=4 \pi \in_{0} \mathrm{R}$
$=4 \pi \in_{0} \frac{3 \mathrm{V}}{\mathrm{~A}}$
$=\frac{12 \pi \in_{0} \mathrm{V}}{\mathrm{~A}}$.