Q131 / 150 · MHT CET 2026 (11 April Shift 2) · Physics
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)
- A$2\pi \epsilon_0 V/A$
- B$4\pi \epsilon_0 V/A$
- C$8\pi \epsilon_0 V/A$
- Correct answerD$12\pi \epsilon_0 V/A$
Solution
The capacitance of a spherical conductor of radius $R$ is given by $C = 4\pi \epsilon_0 R$.
The volume of the sphere is $V = \dfrac{4}{3}\pi R^3$ and its surface area is $A = 4\pi R^2$.
Taking the ratio of volume to surface area:
$\dfrac{V}{A} = \dfrac{\dfrac{4}{3}\pi R^3}{4\pi R^2} = \dfrac{R}{3}$
This gives $R = \dfrac{3V}{A}$.
Substituting the value of $R$ in the expression for capacitance:
$C = 4\pi \epsilon_0 \left(\dfrac{3V}{A}\right) = 12\pi \epsilon_0 \dfrac{V}{A}$
Answer: $12\pi \epsilon_0 V/A$
The volume of the sphere is $V = \dfrac{4}{3}\pi R^3$ and its surface area is $A = 4\pi R^2$.
Taking the ratio of volume to surface area:
$\dfrac{V}{A} = \dfrac{\dfrac{4}{3}\pi R^3}{4\pi R^2} = \dfrac{R}{3}$
This gives $R = \dfrac{3V}{A}$.
Substituting the value of $R$ in the expression for capacitance:
$C = 4\pi \epsilon_0 \left(\dfrac{3V}{A}\right) = 12\pi \epsilon_0 \dfrac{V}{A}$
Answer: $12\pi \epsilon_0 V/A$
Hence, the correct answer is option: 'D'.