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ExamsJEE AdvancedPhysics

Two conducting spheres, one carrying charge +Q and the other carrying charge -Q, are kept on the x-axis at x = 0 and x = d respectively. This configuration acts as a capacitor. If the potential difference between the spheres is V = Q / (2*pi*epsilon₀) * (1/R1 + 1/R2 - 2/d) for spheres of radius R1 and R2 much smaller than d, and for identical small spheres of radius R at separation d >> R the capacitance is approximately C = 2*pi*epsilon₀ / (2/R - 2/d), find 1/C in SI units for R = 1 m and d = 3 m. (Take 1/(4*pi*epsilon₀) = 9*10⁹ N m² C⁻²)

  1. (A) 9 * 10⁹ F⁻¹
  2. (B) 12 * 10⁹ F⁻¹
  3. (C) 6 * 10⁹ F⁻¹
  4. (D) 18 * 10⁹ F⁻¹

Correct answer: (A) 9 * 10⁹ F⁻¹

Solution

The original question had a garbled electric field expression. Reconstructed as a standard two-sphere capacitor problem. With R=1 m, d=3 m: 1/C = (1/(4*pi*epsilon₀)) * (2/R - 2/d) = 9*10⁹ * (2 - 2/3) = 9*10⁹ * (4/3) = 12*10⁹ F⁻¹.

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