Question

A hollow charged metal sphere has radius $r$. If the potential difference between its surface and a point at a distance $3r$ from the centre is $V$, then electric field intensity at distance $3r$ from the centre is

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

Answer 1

The correct option is C $\frac{V}{6r}$

Electric potential at the surface of the sphere, $V_A=\frac{KQ}{r}$
where $K=\frac{1}{4\pi {ϵ}_o}$
Electric potential at distance $3r$ from the centre, $V_B=\frac{KQ}{3r}$
Potential difference, $V=V_A-V_B=\frac{KQ}{r}-\frac{KQ}{3r}$
$V=\frac{2}{3}\frac{KQ}{r}\Rightarrow KQ=\frac{3rV}{2}$
Electric field intensity at $3r$ distance from the centre, $E=\frac{KQ}{(3r{)}^2}=\frac{KQ}{9r^2}=\frac{3}{2}\frac{rV}{9r^2}\therefore E=\frac{V}{6r}$