Question

A solid sphere of radius 20 cm is charged uniformly. The distance from the surface of sphere, where the electrostatic potential is half of that at centre is

Answer 1

Explanation:

$20 \div 3$

always at distance r/3

Answer 2

The distance from the surface of sphere, where the electrostatic potential is half of that at centre is 20/3 cm

Explanation:

For a uniformly charged solid sphere of radius R, the electric potential is given by

$\boxed{V=\frac{kq}{r}}$    For $r>R$

And

$\boxed{V=\frac{kq}{2R^3}(3R^2-r^2)}$    For $r\le R$

Given

The radius of the sphere R = 20 cm

For centre

r = 0

Thus,

Electrostatic potential at the centre

$V_c=\frac{kq}{2R^3}(3R^2-0)$

$\implies V_c=\frac{3}{2}\frac{kQ}{R}$

If the electrostatic potential at a distance r from centre is half of that of centre

Then

$V=\frac{V_c}{2}$

$\implies \frac{kq}{r}=\frac{1}{2}\times\frac{3}{2}\frac{kQ}{R}$

$\implies r=\frac{4}{3}R$

But r is the distance from the centre

Therefore, distance from the surface

$=\frac{4}{3}R-R$

$=\frac{R}{3}$

$=\frac{20}{3}$ cm

Hope this answer is helpful.

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