Question

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

Answer 1

$V_{surface}=\frac{Kq}{r}$

$V_{atdistance3rfromcentre}$$=\frac{Kq}{3r}$

Potential difference $=V=\frac{Kq}{r}-\frac{kq}{3r}$

$=\frac{Kq}{r}.\frac{2}{3}=\frac{2Kq}{3r}$

Electric intensity (at distance $3r$ from the centre$)=$

$\Rightarrow E_{3r}=\frac{Kq}{q{r}^2}$

$\because V=\frac{2}{3}\frac{Kq}{r}$

$\frac{Kq}{r}=\frac{3V}{2}$

$\therefore E_{3r}=\frac{3V}{2\times 9\times r}=\frac{V}{6r}$

$E_{3r}=\frac{V}{6r}$ $[E_{3r}=$ Electric intensity at distance $3r$ from the centre$]$