A charge Q is distributed over two concentric conducting thin spherical shells radii r and R `(R gt 1)` . If the surface charge densities on the two shells are equal, the electric potential at the common centre is:

Options

A. `sqrt(2)/(pi epsilon_0) (Q(R + r))/((R^2 + r^2))`
B. `(1)/(2 pi epsilon_0) (Q(R + r))/((R^2 + r^2))`
C. `(1)/(4 pi epsilon_0) (Q(R + r))/((R^2 + r^2))`
D. `(1)/(pi epsilon_0) (Q(R - r))/((R^2 + r^2))`

Answer

C

Explanation

If `q_1` and `q_2` are the charges on the spheres of radii r and R, respectively, then by conservation of charge.
.
`q_1 + q_2 = Q` …(i)
But `sigma_(1)=sigma_(2)`
`q_(1)/(4pi r^(2))=q_(2)/(4pi R^(2))`
or `q_(1)/q_(2)=r^(2)/R^(2)`
From Eqs. (i) and (ii), we get
`q_1 = (Qr^2)/(r^2 + R^2) and q_2 = (Q R^2)/(r^2 + R^2)`
Hence, potential at common center is
`V = V_1 + V_2`
=`(1)/(4 pi epsilon_0) [q_(1)/r+q_(2)/R]`
=`(1)/(4 pi epsilon_0) [(Q r)/((R^2 + r^2))+ (Q R)/((R^2 + r^2))]`
=`(1)/(4 pi epsilon_0) (Q(R + r))/((R^2 + r^2))`.