Q123 / 150 · MHT CET 2025 (22 Apr Shift 2) · Physics

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)

  • A
    $\frac{2 \pi \epsilon_0 V}{A}$
  • B
    $\frac{8 \pi \epsilon_0 V}{A}$
  • Correct answer
    C
    $\frac{12 \pi \epsilon_0 V}{A}$
  • D
    $\frac{4 \pi \epsilon_0 V}{A}$

Solution

The capacitance of an isolated conducting sphere is $C = 4\pi\epsilon_0 R$, while its volume and surface area are $V = \frac{4}{3}\pi R^3$ and $A = 4\pi R^2$ respectively.

Eliminating the radius via $V/A = \frac{1}{3}R$ yields $R = 3V/A$.

Substituting into the capacitance formula gives $C = 4\pi\epsilon_0 (3V/A) = 12\pi\epsilon_0 V/A$, matching option C.

Hence, the correct answer is option: 'C'.

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