# A thin sphericalal conducting shell of radius r has a charge q. Another charge Q is placed at the centre of the shell. The electrostatic potential at a point p at a distance r / 2 from the centre of the shell is

Arun Kumar IIT Delhi
8 years ago
Hi Ashu,

We will do this with gauss law.
for just a conducting shell field inside is zero since enclosed charge is 0
but here enclosed charge is Q
so
$\\E*4\pi \epsilon_{0} (r/2)^2=Q \\E=Q/4\pi \epsilon_{0} (r/2)^2$

Thanks & Regards

Arun Kumar

IIT Delhi