Q235 / 240 · KCET 2026 · Physics

Consider three point charges $-2Q, Q$ and $-Q$ and three surfaces $S_1, S_2$ and $S_3$ as shown in the figure. Match the entries of List-I with that of List-II.
Figure
List-IList-II
(a) Net flux through $S_1$(i) $\dfrac{-2Q}{\varepsilon_0}$
(b) Net flux through $S_2$(ii) $\dfrac{-Q}{\varepsilon_0}$
(c) Net flux through $S_3$(iii) Zero

  • A
    a - ii, b - i, c – iii
  • B
    a - iii, b - ii, c – i
  • C
    a - i, b - ii, c – iii
  • Correct answer
    D
    a - ii, b - iii, c – i

Solution

According to Gauss's Law, the net electric flux through any closed surface is given by $\Phi = \dfrac{Q_{\text{enclosed}}}{\varepsilon_0}$.

For surface $S_1$, the enclosed charges are $-2Q$ and $Q$.
Net charge enclosed by $S_1 = -2Q + Q = -Q$.
Net flux through $S_1 = \dfrac{-Q}{\varepsilon_0}$.
Thus, (a) matches with (ii).

For surface $S_2$, the enclosed charges are $Q$ and $-Q$.
Net charge enclosed by $S_2 = Q + (-Q) = 0$.
Net flux through $S_2 = 0$.
Thus, (b) matches with (iii).

For surface $S_3$, the enclosed charges are $-2Q$, $Q$, and $-Q$.
Net charge enclosed by $S_3 = -2Q + Q + (-Q) = -2Q$.
Net flux through $S_3 = \dfrac{-2Q}{\varepsilon_0}$.
Thus, (c) matches with (i).

The correct matching is a - ii, b - iii, c - i.

Answer: a - ii, b - iii, c – i

Hence, the correct answer is option: 'D'.

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