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JEE Advanced 2025 · Paper 2, Physics question 7

Six infinitely large and thin non-conducting sheets are fixed in configurations I and II. As shown in the figure, the sheets carry uniform surface charge densities which are indicated in terms of σ0\sigma_0. The separation between any two consecutive sheets is 1 μ\mum. The various regions between the sheets are denoted as 1, 2, 3, 4 and 5. If σ0=9 μC/m2\sigma_0 = 9\ \mu\mathrm{C/m^2}, then which of the following statements is/are correct:

(Take permittivity of free space ϵ0=9×10−12\epsilon_0 = 9\times10^{-12} F/m)

Six parallel vertical sheets, equally spaced 1 μm apart, with regions 1-5 between consecutive sheets. Configuration I charge densities from left to right: +σ₀, −σ₀, +σ₀, −σ₀, +σ₀, −σ₀. Configuration II: +σ₀/2, −σ₀, +σ₀, −σ₀, +σ₀, −σ₀/2.
Choose one or more options
Answer and worked solution

Official answer: A

Field between sheets. In each gap,
E=(charge to the left)−(charge to the right)2ϵ0.E=\frac{(\text{charge to the left})-(\text{charge to the right})}{2\epsilon_0}.
Note σ0ϵ0d=9×10−69×10−12×10−6=1\frac{\sigma_0}{\epsilon_0}d=\frac{9\times10^{-6}}{9\times10^{-12}}\times10^{-6}=1 V.

Configuration I.

  • Gaps 1, 3 and 5 have E=σ0ϵ0E=\frac{\sigma_0}{\epsilon_0}.
  • Gaps 2 and 4 have E=0E=0, so (A) is true.
  • The potential difference is 3×1=33\times1=3 V, so (C) is false.

Configuration II.

  • The field alternates between +σ02ϵ0+\frac{\sigma_0}{2\epsilon_0} and −σ02ϵ0-\frac{\sigma_0}{2\epsilon_0}.
  • Region 3 has σ02ϵ0\frac{\sigma_0}{2\epsilon_0}, not σ0ϵ0\frac{\sigma_0}{\epsilon_0}, so (B) is false.
  • The net potential difference is σ0d2ϵ0=0.5\frac{\sigma_0d}{2\epsilon_0}=0.5 V, not zero, so (D) is false.
Topics:Electric fieldElectric potential
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