MHT CET · Physics · Capacitance
The earth is assumed to be charged conducting sphere having volume \(V\) and surface area \(A\). The capacitance of the earth in free space is
- A \(\frac{2 \pi \epsilon_0 V}{A}\)
- B \(\frac{12 \pi \epsilon_0 V}{A}\)
- C \(\frac{8 \pi \epsilon_0 V}{A}\)
- D \(\frac{4 \pi \epsilon_0 V}{A}\)
Answer & Solution
Correct Answer
(B) \(\frac{12 \pi \epsilon_0 V}{A}\)
Step-by-step Solution
Detailed explanation
If we assume shape of earth be solid sphere of radius \(R\).
The ratio of earth volume to surface area: \(\frac{V}{A}=\frac{\left(\frac{4}{3} \pi R^3\right)}{4 \pi R^2}=\frac{R}{3}\) or, \(R=\frac{3 V}{A}\)
The capacitance of earth is,
\(C_E=4 \pi \epsilon_0 R=4 \pi \epsilon_0\left(\frac{3 V}{A}\right)=\frac{12 \pi \epsilon_0 V}{A}\)
The ratio of earth volume to surface area: \(\frac{V}{A}=\frac{\left(\frac{4}{3} \pi R^3\right)}{4 \pi R^2}=\frac{R}{3}\) or, \(R=\frac{3 V}{A}\)
The capacitance of earth is,
\(C_E=4 \pi \epsilon_0 R=4 \pi \epsilon_0\left(\frac{3 V}{A}\right)=\frac{12 \pi \epsilon_0 V}{A}\)
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