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MHT CET 2026 (19 May, Shift 1) · Capacitance of a Conductor

Capacitance of a Conductor question from MHT CET 2026 (19 May, Shift 1)

Q2MHT CET 2026 · 19 MAY · SHIFT 1Easy

Capacitance of a Conductor

Consider the earth to be a sphere, having charge QQ. Let R,VR, V and AA be its radius, volume and area respectively. The capacitance of the earth is [ε0\varepsilon_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) 12πε0VA12\pi\varepsilon_0\frac{V}{A}.
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