In an electric field due to charge $Q$, a charge $q$ moves from point $A$ to $B$ as shown in the figure. The…
- $\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\mathrm{r}^2}$
- $\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\mathrm{r}^2} \frac{\pi}{6}$
- $\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\mathrm{r}}$
- zero
Solution
Electric field is perpendicular to the displacement. $\overrightarrow{\mathrm{F}}=\mathrm{q} \overrightarrow{\mathrm{E}}$ $\therefore \quad \vec{F}$ is also perpendicular to the curved surface $\begin{aligned} \mathrm{W} & =\int \mathrm{dW} \\ & =\int \overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{ds}}=\mathrm{F} \cdot \mathrm{ds} \cos 90^{\circ} \\ \therefore \quad \mathrm{W} & =0 \end{aligned}$
Asked in: MHT CET 2024 (15 May Shift 2)