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

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$\frac{12 \pi \epsilon_0 V}{A}$

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(B) Assume the Earth is a solid sphere of radius $R$.The volume $V$ of a sphere is given by $V = \frac{4}{3} \pi R^3$ and the surface area $A$ is given by $A = 4 \pi R^2$.The ratio of volume to surface area is $\frac{V}{A} = \frac{\frac{4}{3} \pi R^3}{4 \pi R^2} = \frac{R}{3}$.From this,we can express the radius $R$ as $R = \frac{3V}{A}$.The capacitance $C$ of an isolated spherical conductor of radius $R$ is given by $C = 4 \pi \epsilon_0 R$.Substituting the value of $R$ in terms of $V$ and $A$,we get $C = 4 \pi \epsilon_0 \left( \frac{3V}{A} \right) = \frac{12 \pi \epsilon_0 V}{A}$.

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

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