MHT CET · Physics · Electrostatics
In the electric field due to a charge ' \(Q\) ', a charge ' \(q\) ' moves from point A to B. The work done is ( \(\varepsilon_0=\) permittivity of vacuum)

- A \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\mathbf{r}^2}\)
- B \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\mathrm{r}} \times \frac{\pi}{6}\)
- C \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\mathrm{r}}\)
- D Zero
Answer & Solution
Correct Answer
(D) Zero
Step-by-step Solution
Detailed explanation
The points A and B are equipotential surfaces as both are at the same distance from the charge Q. Therefore, the work done is zero.
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