WBJEE · Physics · Electrostatics
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
- 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 \(\frac{\mathrm{qQ}}{4 \pi \in_0 \mathrm{~L}}, 0\)
Answer & Solution
Correct Answer
(A) \(\frac{-\mathrm{Qq}}{6 \pi \epsilon_0 \mathrm{~L}}, \frac{-\mathrm{Qq}}{6 \pi \epsilon_0 \mathrm{~L}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { } W_{C S D}=Q\left(V_{D A}-V_{C A}\right)=\frac{Q q}{4 \pi \epsilon_0 L}\left[\frac{1}{3}-1\right]=\frac{-2 Q q}{12 \pi \epsilon_0 L}=\frac{-Q q}{6 \pi \epsilon_0 L}=W_{C B D} \\ & W_{C B D}=W_{C S D} \text { path independent }\end{aligned}\)
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