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 field intensity at a distance ‘3r’ is ______.

Options

• `v/(2 r)`

• `v/(3 r)`

• `v/(6 r)`

• `v/(4 r)`

Answer 1

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 field intensity at a distance ‘3r’ is `bbunderline(v/(6 r))`.

Explanation:

Given: Potential difference between r and 3r:

V(r) − V(3r) = v

`(kQ)/r - (kQ)/(3r)` = v

`(kQ)/r (1 - 1/3)` = v

`(kQ)/r * 2/3` = v

`(kQ)/r = (3v)/2`    ...(i)

Electric field at 3r:

E = `(kQ)/((3r)^2)`

= `(kQ)/(9r^2)`    ...[From equation (i)]

= `1/(9r^2) * ((3v)/2 * r)`

= `(3v)/(18 r)`

= `v/(6r)`