The earth has volume V and surface area A, then its capacitance would…

The correct answer is Option (3) → 12πε₀V/A

Treat the earth as an isolated conducting sphere of radius R, whose capacitance is \( 4\pi\varepsilon_0 R \). Express R through the given volume and surface area.

    \( \displaystyle V = \frac{4}{3}\pi R^{3}, \quad A = 4\pi R^{2} \)

    \( \displaystyle \frac{V}{A} = \frac{\frac{4}{3}\pi R^{3}}{4\pi R^{2}} = \frac{R}{3} \)

    \( \displaystyle R = \frac{3V}{A} \)

    \( \displaystyle C = 4\pi\varepsilon_0 R = 4\pi\varepsilon_0 \times \frac{3V}{A} = \frac{12\pi\varepsilon_0 V}{A} \)

Shortcut: Check the units: capacitance must be \( \varepsilon_0 \times \) length. V/A has the unit of length, but A/V has the unit of 1/length, which rules out Options (1) and (4). Of the remaining two, only a factor of 3 turns V/A into the radius.