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(A) Assuming the Earth is a solid sphere of radius $R$. The volume of the sphere is $V = \frac{4}{3} \pi R^{3}$. The surface area of the sphere is $A

Vedclass · Accepted Answer

$12 \pi \epsilon_{0} \frac{V}{A}$

Vedclass · Answer

(A) Assuming the Earth is a solid sphere of radius $R$. The volume of the sphere is $V = \frac{4}{3} \pi R^{3}$. The surface area of the sphere is $A = 4 \pi R^{2}$. Dividing the volume by the surface area,we get: $\frac{V}{A} = \frac{\frac{4}{3} \pi R^{3}}{4 \pi R^{2}} = \frac{R}{3}$. From this,we find the radius $R = 3 \frac{V}{A}$. The capacitance $C$ of an isolated spherical conductor is given by $C = 4 \pi \epsilon_{0} R$. Substituting the value of $R$,we get: $C = 4 \pi \epsilon_{0} \left( 3 \frac{V}{A} \right) = 12 \pi \epsilon_{0} \frac{V}{A}$.

The Earth is assumed to be a charged conducting sphere having volume $V$ and surface area $A$. The capacitance of the Earth in free space is ($\epsilon_{0} =$ permittivity of free space).

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