#
Solve this:

Q. Consider a sphere of radius R having charge q uniformly distributed inside it. At what minimum distance from its surface the electric potential is half of the electric potential at its centre?

$\left(1\right)\mathrm{R}\left(2\right)\frac{\mathrm{R}}{2}\phantom{\rule{0ex}{0ex}}\left(3\right)\frac{4\mathrm{R}}{3}\left(4\right)\frac{\mathrm{R}}{3}$

we know potential outside the sphere is V1 = kQ/x where x is the distance from the center x>r

and potential inside the sphere V1 = kQ/2r [3-x2/r2]

for center put x=0 Vo = 3kQ/2r

condition given is V1 = Vo/2

kQ/x = (3kQ/2r)/2

x=4r/3

distance from the surface is =x-r

= 4r/3 -r =r/3

Regards

**
**