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# A charge is distributed over two concentric hollow spheres of radii r and R where R>r , such that the surface densities of charges are equal .what is potential at their common centre

Zachariah Abraham
31 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:
$k(Q)(r+R)/(r^2 + R^2).$

Rishi Sharma
one year ago
Hello students,
The solution of the above problem is in the attached file.
I hope the solution will solve all your doubts.
Thank You,
All the Best for the Exams.