Question

A charge 'q' is distributed over two concentric hollow co
ducting sphere of radii r and R(>r) such that their surfa
charge densitites are equal. The potential at their comm
centre is

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Answer 1

Explanation:σ(4πr2+4πR2)=Q  

∴σ=Q4π(r2+R2)  

Potential at centre  

= potential due to A+potential due to B  

=σrε0+σRε0  

=σε0(r+R)  

(Q4πε0)(r+Rr2+R2)

Answer 2

Answer:

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).

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