Q141 / 155 · WBJEE 2025 · Physics

Two charges $+q$ and $-q$ are placed at points $A$ and $B$ respectively which are at a distance $2 L$ apart. $C$ is the mid point of $A$ and $B$. The workdone in moving a charge $+Q$ along the semicircle $\operatorname{CSD}\left(W_1\right)$ and along the line $\operatorname{CBD}\left(W_2\right)$ are Figure

  • Correct answer
    A
    $\frac{-\mathrm{Qq}}{6 \pi \epsilon_0 \mathrm{~L}}, \frac{-\mathrm{Qq}}{6 \pi \epsilon_0 \mathrm{~L}}$
  • B
    $\frac{\mathrm{qQ}}{4 \pi \epsilon_0 \mathrm{~L}}, \frac{\mathrm{qQ}}{4 \pi \epsilon_0 \mathrm{~L}}$
  • C
    $\frac{-Q q}{6 \pi \epsilon_0 L}, \frac{-Q q}{12 \pi \epsilon_0 L}$
  • D
    $\frac{\mathrm{qQ}}{4 \pi \in_0 \mathrm{~L}}, 0$

Solution

$\begin{aligned} & \text { } W_{C S D}=Q\left(V_{D A}-V_{C A}\right)=\frac{Q q}{4 \pi \epsilon_0 L}\left[\frac{1}{3}-1\right]=\frac{-2 Q q}{12 \pi \epsilon_0 L}=\frac{-Q q}{6 \pi \epsilon_0 L}=W_{C B D} \\ & W_{C B D}=W_{C S D} \text { path independent }\end{aligned}$

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

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