Question:medium

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 (\( \varepsilon_0 = \) permittivity of free space)}

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Isolated Sphere: $C = 4\pi\varepsilon_0 R$.
Updated On: May 7, 2026
  • \( \frac{2\pi\varepsilon_0 V}{A} \)
  • \( \frac{8\pi\varepsilon_0 V}{A} \)
  • \( \frac{12\pi\varepsilon_0 V}{A} \)
  • \( \frac{4\pi\varepsilon_0 V}{A} \)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
An isolated conducting sphere has a capacitance that depends only on its geometry (its radius) and the permittivity of the surrounding medium.
We need to express this capacitance in terms of the sphere's volume \( V \) and surface area \( A \).
Step 2: Key Formula or Approach:
The capacitance \( C \) of an isolated spherical conductor of radius \( R \) is: \[ C = 4\pi\varepsilon_0 R \] The volume \( V \) of a sphere is: \[ V = \frac{4}{3}\pi R^3 \] The surface area \( A \) of a sphere is: \[ A = 4\pi R^2 \] Step 3: Detailed Explanation:
We want to replace the radius \( R \) in the capacitance formula with an expression involving \( V \) and \( A \).
Let's find the ratio of volume to surface area: \[ \frac{V}{A} = \frac{\frac{4}{3}\pi R^3}{4\pi R^2} \] Simplify the expression by canceling \( 4\pi \) and \( R^2 \): \[ \frac{V}{A} = \frac{R}{3} \] Now, rearrange this equation to solve for the radius \( R \): \[ R = \frac{3V}{A} \] Substitute this expression for \( R \) into the formula for capacitance: \[ C = 4\pi\varepsilon_0 \left( \frac{3V}{A} \right) \] Multiply the constants together: \[ C = \frac{12\pi\varepsilon_0 V}{A} \] This matches option (C).
Step 4: Final Answer:
The capacitance of the earth is \( \frac{12\pi\varepsilon_0 V}{A} \).
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