Question

Three concentric spherical shells have radii $a,\,b$ and $c$ $(a(A). ${{V}_{C}}={{V}_{B}}={{V}_{A}}$
(B). ${{V}_{C}}={{V}_{B}}\ne {{V}_{A}}$
 (C). ${{V}_{C}}\ne {{V}_{B}}={{V}_{A}}$
(D). ${{V}_{C}}\ne {{V}_{B}}\ne {{V}_{A}}$

Answer 1

Hint: The potential due to a spherical shell is the work done to move a charge from one point to the other. It depends on the charge permittivity of medium and distance from the centre. The potential on each sphere will be due to their own potential and the potential of other spheres also.

Formula used:
$E=\dfrac{q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}$
$\sigma =\dfrac{q}{4\pi {{r}^{2}}}$

Complete answer:
The electric field is defined as the work done to bring a unit charge from infinity to a point in the field. Its SI unit is $Vm$.
The electric field due to a spherical shell is given by-
$E=\dfrac{q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}$ - (1)
Here, $E$ is the electric field
$q$ is the charge on the spherical shell
${{\varepsilon }_{0}}$ is the permittivity of free space
$r$ is the distance from the centre of the shell
Given three concentric spheres having radii a, b, c such that $(a

Charge density is the charge per unit area. Therefore for a spherical shell,
$\sigma =\dfrac{q}{4\pi {{r}^{2}}}$
$\Rightarrow q=\sigma 4\pi {{r}^{2}}$ - (2)
Therefore, the electric field is independent of the radius.
The electric potential is defined as the work done to move a charge between two points in an electric field. For a conducting shell the potential is given by-
$V=\dfrac{q}{4\pi {{\varepsilon }_{0}}r}$
For the spherical shell of radius a, the potential will be-
$\begin{align}
  & {{V}_{a}}=\dfrac{q}{4\pi {{\varepsilon }_{0}}a}+\dfrac{q}{4\pi {{\varepsilon }_{0}}b}+\dfrac{q}{4\pi {{\varepsilon }_{0}}c} \\
 & \Rightarrow {{V}_{a}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\left( \dfrac{q}{a}+\dfrac{q'}{b}+\dfrac{q''}{c} \right) \\
\end{align}$
Substituting charge in terms of charge density from eq (2), we get,
  $\begin{align}
  & {{V}_{a}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\left( \dfrac{\sigma 4\pi {{a}^{2}}}{a}+\dfrac{-\sigma 4\pi {{b}^{2}}}{b}+\dfrac{\sigma 4\pi {{c}^{2}}}{c} \right) \\
 & \Rightarrow {{V}_{a}}=\dfrac{\sigma }{{{\varepsilon }_{0}}}\left( a-b+c \right) \\
\end{align}$
For spherical shell for radius b and charge density$-\sigma $, the potential will be
$\begin{align}
  & {{V}_{b}}=\dfrac{q}{4\pi {{\varepsilon }_{0}}b}+\dfrac{q}{4\pi {{\varepsilon }_{0}}b}+\dfrac{q}{4\pi {{\varepsilon }_{0}}c} \\
 & \Rightarrow {{V}_{b}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\left( \dfrac{q}{b}+\dfrac{q'}{b}+\dfrac{q''}{c} \right) \\
\end{align}$
Substituting charge in terms of charge density, we get,
$\begin{align}
  & {{V}_{b}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\left( \dfrac{\sigma 4\pi {{a}^{2}}}{b}+\dfrac{-\sigma 4\pi {{b}^{2}}}{b}+\dfrac{\sigma 4\pi {{c}^{2}}}{c} \right) \\
 & \Rightarrow {{V}_{b}}=\dfrac{\sigma }{{{\varepsilon }_{0}}}\left( \dfrac{{{a}^{2}}}{b}-b+c \right) \\
\end{align}$
For spherical shell with radius c and charge density $+\sigma $ is-
$\begin{align}
  & {{V}_{c}}=\dfrac{q}{4\pi {{\varepsilon }_{0}}c}+\dfrac{q}{4\pi {{\varepsilon }_{0}}c}+\dfrac{q}{4\pi {{\varepsilon }_{0}}c} \\
 & \Rightarrow {{V}_{c}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\left( \dfrac{q}{c}+\dfrac{q'}{c}+\dfrac{q''}{c} \right) \\
\end{align}$
Substituting charge in terms of charge density, we get,
$\begin{align}
  & {{V}_{c}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\left( \dfrac{\sigma 4\pi {{a}^{2}}}{c}+\dfrac{-\sigma 4\pi {{b}^{2}}}{c}+\dfrac{\sigma 4\pi {{c}^{2}}}{c} \right) \\
 & \Rightarrow {{V}_{c}}=\dfrac{\sigma }{{{\varepsilon }_{0}}}\left( \dfrac{{{a}^{2}}}{c}-\dfrac{{{b}^{2}}}{c}+c \right) \\
\end{align}$
Therefore, the potential of the spherical shell of radius a is $\dfrac{\sigma }{{{\varepsilon }_{0}}}\left( a-b+c \right)$, the potential of spherical shell of radius b is $\dfrac{\sigma }{{{\varepsilon }_{0}}}\left( \dfrac{{{a}^{2}}}{b}-b+c \right)$ and the potential of shell with radius c is $\dfrac{\sigma }{{{\varepsilon }_{0}}}\left( \dfrac{{{a}^{2}}}{c}-\dfrac{{{b}^{2}}}{c}+c \right)$, none of the potentials are equal.

Hence, the correct option is (D).

Note:
There is no charge inside a conductor; all the charge is concentrated on the surface. The potential inside the sphere is constant and is equal to the potential on the surface. Charge density can also be expressed as linear charge density and volume charge density.