Question

A charge Q is distributed over two concentric hollow spheres of radii $r$ and $R$ $\left( R>r \right)$ such that their surface densities are equal. The potential at the common centre is:
A. $\dfrac{\sqrt{2}}{\pi {{\varepsilon }_{0}}}\dfrac{Q\left( R+r \right)}{\left( {{R}^{2}}+{{r}^{2}} \right)}$
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. $\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{Q\left( R+r \right)}{\left( {{R}^{2}}+{{r}^{2}} \right)}$

Answer 1

Surface charge density which is denoted as $\sigma$ is a quantity of charge per unit area. It is measured in coulombs divided by square meter $c{{m}^{-2}}$, on a two dimensional surface at any of the points on a surface charge distribution.

Complete step by step answer:
First of all let us check what a surface charge density actually means. Surface charge density which is denoted as is a quantity of charge per unit area. It is measured in coulombs divided by square meters, on a two dimensional surface at any of the points on a surface charge distribution. Potential is the amount of work done in order to move a unit charge to a specific point from a reference point inside an electric field without having an acceleration. It is given by the equation
$V=\dfrac{Q\left( R+r \right)}{4\pi {{\varepsilon }_{0}}\left( {{R}^{2}}+{{r}^{2}} \right)}$
Where $V$ is the potential, $Q$ is the charge, $r$ and $R$ are the radii of concentric hollow spheres.
Suppose the charge on the surface of the sphere of radius $R$ be ${{q}_{A}}$ and that of sphere with radius $r$ be ${{q}_{B}}$,
As we all know their surface charge densities are equal,
Therefore we can write that
$\dfrac{{{q}_{A}}}{{{R}^{2}}}=\dfrac{{{q}_{B}}}{{{r}^{2}}}$
And also we know that,
${{q}_{A}}+{{q}_{B}}=Q$
After solving these equations, we can write that
${{q}_{A}}=\dfrac{Q{{R}^{2}}}{{{R}^{2}}+{{r}^{2}}}$
${{q}_{A}}=\dfrac{Q{{R}^{2}}}{{{R}^{2}}+{{r}^{2}}}$
The potential at the centre will be
$V=\dfrac{k{{q}_{A}}}{R}+\dfrac{k{{q}_{B}}}{r}$
$V=\dfrac{kQ\left( R+r \right)}{4\pi {{\varepsilon }_{0}}\left( {{R}^{2}}+{{r}^{2}} \right)}$
As we know $k=\dfrac{1}{4\pi {{\varepsilon }_{0}}}$
$V=\dfrac{Q\left( R+r \right)}{4\pi {{\varepsilon }_{0}}\left( {{R}^{2}}+{{r}^{2}} \right)}$

Therefore the correct answer is option D.

Note:
Potential is the amount of work done in order to move a unit charge to a specific point from a reference point inside an electric field without having an acceleration. It is a scalar quantity as it depends on the magnitude only. ​The dimensional formula for electric potential is $M{{L}^{2}}{{T}^{-3}}{{I}^{-1}}$.