In the electric field due to a charge ' $Q$ ', a charge ' $q$ ' moves from point A to B. The work done is (…

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)
  1. $\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\mathbf{r}^2}$
  2. $\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\mathrm{r}} \times \frac{\pi}{6}$
  3. $\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\mathrm{r}}$
  4. Zero

Solution

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.

Asked in: MHT CET 2023 (13 May Shift 2)

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