# a q charge is distributed over two concentric spherical shells of radius r and R (R>r) and having same surface charge densities . find the potential at the common centre of the shell

Arun
25757 Points
4 years ago

By superposition princpiple, potential at the common centre is equal to algebraic sum of potentials at centre due to each sphere.

If we want the potential of a sphere, we need the radius (given) and the charge on it (which is what we should find now).
If the total charge is Q, then let’s assume charge of small sphere si q1, and large sphere is q2.
Thus $Q = q1 + q2$

It is given that the surface charge density is the same, thus:
$(q1)/(4*pi*r^2) = (q2)/(4*pi*R^2).$
Therefore,
$q1 = (r^2)(q2)/(R^2)$

But $q1 + q2 = Q,$
therefore,
$q2 = Q(R^2)/(r^2 + R^2),$
and similarly (from the same equation,
$q1 = Q(r^2)/(r^2 + R^2).$
Potential at common centre is now given as:
$k(q1)/r + k(q2)/R.$

Substituting previously found values, this becomes:
$k(Q)(r+R)/(r^2 + R^2).$

Regards
Arun
Khimraj
3007 Points
4 years ago
charge density=q/4(3.14)r=Q/4(3.14)R.
ie, q=Q*r*r/R*R.
potential at center of shell=kq/r+kQ/R
kQ/R(r/R+1)
Q=charge density*4(3.14)R*R
on simpilfying,
V=charge density*(r+R)/epsilon