Question

A charge Q is distributed over two concentric hollow spheres at radii r and R $\left( {R > r} \right)$ such that their surface densities are equal. The potential at the common center is:

A.)$\dfrac{{\sqrt 2 Q\left( {R + r} \right)}}{{\pi {\varepsilon _0}\left( {{R^2} + {r^2}} \right)}}$
B.)$\dfrac{1}{{2\pi {\varepsilon _0}}}\dfrac{{Q\left( {R + r} \right)}}{{\left( {{R^2} + {r^2}} \right)}}$
C.)$\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{Q\left( {R + r} \right)}}{{\left( {{R^2} + {r^2}} \right)}}$
D.)$\dfrac{1}{{\pi {\varepsilon _0}}}\dfrac{{Q\left( {R - r} \right)}}{{\left( {{R^2} + {r^2}} \right)}}$

Answer 1

Hint- In order to solve this question, we will use the concept that potential inside the hollow sphere is constant. Therefore the potential inside the hollow sphere will be the same as the potential on the surface of the hollow sphere.

Complete step by step answer:
If the total charge is Q, then let’s assume charge of small sphere is ${q_1}$ and large sphere is ${q_2}$
Thus \[Q{\text{ }} = {\text{ }}{q_1}{\text{ }} + {\text{ }}{q_2}\]
The potential at the center is given as
$V = \left( {\dfrac{{k{q_1}}}{r}} \right) + \left( {\dfrac{{k{q_2}}}{R}} \right)$
It is given that the surface charge density is the same, thus:
$\sigma = \dfrac{{({q_1})}}{{(4\pi {r^2})}} = \dfrac{{{q_2}}}{{4\pi {R^2}}}$
Where $\sigma $ is surface charge density.
Therefore,
${q_1} = \dfrac{{{q_2}{r^2}}}{{{R^2}}}$
But \[Q{\text{ }} = {\text{ }}{q_1}{\text{ }} + {\text{ }}{q_2}\]
Therefore,
${q_2} = \dfrac{{Q({R^2})}}{{({r^2} + {R^2})}}$
And similarly (from the same equation)
${q_1} = \dfrac{{Q({r^2})}}{{({r^2} + {R^2})}}$
Potential at common center is now given as:
\[V = \dfrac{{k{q_1}}}{r} + \dfrac{{k{q_2}}}{R}\]
Substituting previously found values, this becomes:
$V = \dfrac{{kQ\left( {r + R} \right)}}{{{r^2} + {R^2}}}$
Hence correct option is B
Additional Information- According to Gauss law the field or flux inside a closed surface is equal to the net charge enclosed within that volume divided by the permittivity of the medium.
As in the case of a hollow sphere, the electric field inside the hollow sphere is zero because no charge resides inside the hollow sphere. As we know the charge inside the hollow sphere accumulates on the surface of the hollow sphere.

Note- As we know that potential is a scalar quantity and the potential due to two charges is the sum of the potential difference at that point individually. In the above question we first calculated the relation between charges on the hollow sphere and the total charge and then substituted in the potential of the sphere due to two charges.