Question 711: Electrostatics & Capacitance

Question

If a conducting sphere of radius $$\mathrm{R}$$ is charged. Then the electric field at a distance $$\mathrm{r}(\mathrm{r} > \mathrm{R})$$ from the centre of the sphere would be, $$(\mathrm{V}=$$ potential on the surface of the sphere)

Accepted Answer

$$
\begin{aligned}
& 	herefore V=\frac{K Q}{R} \
& E=\frac{K Q}{r^2} \
& E=\frac{V R}{r^2}
\end{aligned}
$$

Answer

$$\frac{r V}{R^2}$$

Answer

$$\frac{R^2 V}{r^3}$$

Answer

$$\frac{R V}{r^2}$$

Answer

$$\frac{\mathrm{V}}{\mathrm{r}}$$

Electrostatics & Capacitance

If a conducting sphere of radius $\mathrm{R} is charged. Then the electric field at a distance \mathrm{r}(\mathrm{r} > \mathrm{R}) from the centre of the sphere would be, (\mathrm{V}=$ potential on the surface of the sphere)

Choose the correct answer: