Question 800: Electrostats & Capacitance

Question

In an electric field due to charge $Q$, a charge $q$ moves from point A to B as shown in the figure. The work done is ( $\varepsilon_0=$ permittivity of free space)

Accepted Answer

Electric field is perpendicular to the displacement.
$$\overrightarrow{\mathrm{F}}=\mathrm{q} \overrightarrow{\mathrm{E}}$$
$	herefore \quad \overrightarrow{\mathrm{F}}$ is also perpendicular to the curved surface
$$\begin{aligned}
\mathrm{W} & =\int \mathrm{dW} \
& =\int \overrightarrow{\mathrm{F}} \cdot \overline{\mathrm{ds}}=\mathrm{F} \cdot \mathrm{ds} \cos 90^{\circ} \
	herefore \quad \mathrm{W} & =0
\end{aligned}$$

Answer

$\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\mathrm{r}^2}$

Answer

$\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\mathrm{r}^2} \frac{\pi}{6}$

Answer

$\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\mathrm{r}}$

Answer

zero

Electrostats & Capacitance

In an electric field due to charge Q, a charge q moves from point A to B as shown in the figure. The work done is ( \varepsilon_0= permittivity of free space)

Choose the correct answer: