Question

If radius of a hollow metallic sphere is ${}^{\prime }{R}^{\prime }$. If the $P.D$ between it's surface and a point at a distance $3R$ from it's centre is $V$ then the electric field intensity at a distance $3R$ from it's centre is

• A. $\frac{V}{2R}$
• B. $\frac{V}{3R}$
• C. $\frac{V}{4R}$
• D. $\frac{V}{6R}$

Answer 1

The correct option is D $\frac{V}{6R}$

The charge $Q$ on a hollow metal sphere is uniformly distributed on it surface. This means that the potential and electric field outside the sphere are the same as these quantities for a point charge $Q$ placed at the centre of the sphere.

Thus, the potential difference between the surface and the point at a distance $3R$ is

$V=\frac{1}{4\pi {ϵ}_0}(\frac{Q}{r}-\frac{Q}{3r})$

$=\frac{2}{3}\frac{1}{4\pi {ϵ}_0}\frac{Q}{r}$

The electric field at this point is

$E=\frac{1}{4\pi {ϵ}_0}\frac{Q}{{\left(3r\right)}^2}$

$=\frac{1}{9}\frac{1}{4\pi {ϵ}_0}\frac{Q}{r^2}$

$=\frac{1}{9}\frac{\frac{2}{3}V}{r}=\frac{1}{6}\frac{V}{r}$