Question

A hollow charged metal sphere has radius r if the potential difference between its surface

Answer 1

Explanation:

Potential at the surface of hollow sphere , V = $\frac{kq}{r}$ , where q is charge distributed on the surface of hollow sphere.

Potential at the surface - potential at 3r distance from the centre = v

or, $\frac{kq}{r}$ - $\frac{kq}{3r}$ = v

or, $\frac{2kq}{3r}$ = v  

or, q = $\frac{3rv}{2k}$ .....(1)

Now electric field intensity at a distance 3r from the centre is

E = $\frac{kq}{3r}$ square

From equation (1),

E = k(3rv/2k)/9r²  

= 3rv/18r²  

= $\frac{v}{6r}$

Hence, the answer is v/6r.