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MHT CET 2026 (11 Apr, Shift 2) · Capacitance of a Conductor

Capacitance of a Conductor question from MHT CET 2026 (11 Apr, Shift 2)

Q3MHT CET 2026 · 11 APR · SHIFT 2Easy

Capacitance of a Conductor

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 (ϵ0\epsilon_0 = permittivity of free space)

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Answer: (d)

  1. Capacitance of a sphere: C=4πε0RC = 4\pi\varepsilon_0 R.
  2. VA=43πR34πR2=R3⇒R=3VA\frac{V}{A} = \frac{\frac{4}{3}\pi R^3}{4\pi R^2} = \frac{R}{3} \Rightarrow R = \frac{3V}{A}
  3. C=4πε0⋅3VAC = 4\pi\varepsilon_0 \cdot \frac{3V}{A}
  4. So the answer is (d) 12πϵ0V/A12\pi\epsilon_0 V/A.
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