Question

A conducting sphere of radius $R$ is charged to a potential of $V$ volt. Then the electric field at a distance $r( > R)$ from the centre of the sphere would be

• A. $\frac{R V}{r^{2}}$
• B. $\frac{V}{r}$
• C. $\frac{r V}{R^{2}}$
• D. $\frac{R^{2} V}{r^{3}}$

Answer 1

Answer: (A)

As $V=\frac{q}{4 \pi \varepsilon_{0} R}$ or
$\frac{q}{4 \pi \varepsilon_{0}}=V R \dots$(i)
$E=\frac{q}{4 \pi \varepsilon_{0} r^{2}}$ For $r > R $
$ \therefore E=\frac{V R}{r^{2}}$ (Using (i))