JEE Advanced 2025 Paper 2 Q07 Physics Electrostatics & Circuits Gauss's Law Hard

JEE Advanced 2025 Paper 2 · Q07 · Gauss's Law

Six infinitely large and thin non-conducting sheets are fixed in configurations I and II as shown in the figure above. The sheets carry uniform surface charge densities expressed in terms of $\sigma_0$. The separation between any two consecutive sheets is $1\,\mu$m. The various regions between the sheets are denoted by 1, 2, 3, 4 and 5. If $\sigma_0 = 9\,\mu$C/m$^2$, then which of the following statements is correct? (Take permittivity of free space $\epsilon_0 = 9 \times 10^{-12}$ F/m.)

\textit{Note: officially Section 2 'one or more correct'; the official key marks only one correct, so we render this as single-correct.}

  1. A. In region 4 of configuration I, the magnitude of the electric field is zero.
  2. B. In region 3 of configuration II, the magnitude of the electric field is $\sigma_0/\epsilon_0$.
  3. C. Potential difference between the first and the last sheets of configuration I is 5 V.
  4. D. Potential difference between the first and the last sheets of configuration II is zero.
Reveal answer + step-by-step solution

Correct answer:A

Solution

Each infinite sheet of density $\sigma$ contributes $|E|=\sigma/(2\epsilon_0)$ on each side, directed away from $+\sigma$ and toward $-\sigma$. Superpose all six sheets in each region.

(A) Config I = $+\sigma_0,-\sigma_0,+\sigma_0,-\sigma_0,+\sigma_0,-\sigma_0$. In region 4 the three sheets to the left ($+,-,+,-$ net $0$ contribution to the right of the 4th sheet, after careful accounting of directions) and the two sheets to the right cancel exactly: net $E=0$. TRUE.

(B) Config II $=+\sigma_0/2,-\sigma_0,+\sigma_0,-\sigma_0,+\sigma_0,-\sigma_0/2$. Sum in region 3 yields $|E|=\sigma_0/(2\epsilon_0)$, not $\sigma_0/\epsilon_0$. FALSE.

(C) Across config I the 5 gap contributions sum to $V_{15}\neq 5$ V (with $d=1\,\mu$m, $\sigma_0/\epsilon_0=10^6$ V/m gives 1 V/gap; signs alternate, giving net $\neq 5$). FALSE.

(D) Outside config II, symmetric outer $\pm\sigma_0/2$ make the exterior field zero, but the interior inter-sheet fields all align, giving a non-zero $V_{15}$. FALSE.

Only (A) is correct.

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