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# Three concentric metal shells A, B and C of respectively radii $a$, $b$ & $c$($a < b < c$) have surface charge densities $+ \sigma$, $- \sigma$ and $+ \sigma$ respectively. The potential of shell B is?A) $\dfrac{\sigma }{{{\varepsilon _0}}}\left[ {\dfrac{{{b^2} - {c^2}}}{b} + a} \right]$B) $\dfrac{\sigma }{{{\varepsilon _0}}}\left[ {\dfrac{{{b^2} - {c^2}}}{c} + a} \right]$C) $\dfrac{\sigma }{{{\varepsilon _0}}}\left[ {\dfrac{{{a^2} - {b^2}}}{a} + c} \right]$D) $\dfrac{\sigma }{{{\varepsilon _0}}}\left[ {\dfrac{{{a^2} - {b^2}}}{b} + c} \right]$

Last updated date: 13th Jun 2024
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Hint: Remember that, potential of a shell will be affected by the charge enclosed in the nearby shells. If the distance between them is smaller when compared to the radius, take radius as the distance, If not take distance between them.

Complete step by step solution:
Let’s define all the data given in the question:
Radii of shell A= $a$
Radii of shell B= $b$
Radii of shell C= $c$
Surface charge density of shell A= $+ \sigma$
Surface charge density of shell B= $- \sigma$
Surface charge density of shell C= $+ \sigma$
We need to find the potential of shell B.
Potential of shell B will be affected by the charge enclosed in all the three shells, so we get,
${V_B} = \dfrac{{K{q_A}}}{b} + \dfrac{{K{q_B}}}{b} + \dfrac{{K{q_C}}}{c}$
${q_A}$ = the charges enclosed in shell A.
${q_B}$ = the charges enclosed in shell B.
${q_C}$ = the charges enclosed in shell C.
The charges enclosed in the shell A,
${q_A} = \sigma (4\pi {a^2})$
The charges enclosed in the shell B,
${q_B} = - \sigma (4\pi {b^2})$
The charges enclosed in the shell C,
${q_C} = \sigma (4\pi {c^2})$
K is a constant and which is given by, $K = \dfrac{1}{{4\pi {\varepsilon _0}}}$
(${\varepsilon _0}$ is the permittivity in vacuum)
Apply these values to the equation for potential of B, we get,
$\Rightarrow {V_B} = \dfrac{{\sigma 4\pi }}{{4\pi {\varepsilon _0}}}\left[ {\dfrac{{{a^2}}}{b} - \dfrac{{{b^2}}}{b} + \dfrac{{{c^2}}}{c}} \right]$
Some of the terms gets cancelled:
$\Rightarrow {V_B} = \dfrac{\sigma }{{{\varepsilon _0}}}\left[ {\dfrac{{{a^2}}}{b} - b + c} \right]$
$\Rightarrow {V_B} = \dfrac{\sigma }{{{\varepsilon _0}}}\left[ {\dfrac{{{a^2} - {b^2}}}{b} + c} \right]$
No we get the value of the potential of the shell B;
So the final answer is option (D). $\dfrac{\sigma }{{{\varepsilon _0}}}\left[ {\dfrac{{{a^2} - {b^2}}}{b} + c} \right]$.

Note: The electric potential difference between the inner and outer surface of different states of the object is described as the surface charge. The surface charge density describes the whole amount of charge per unit amount of the area and it will be there only in conducting surfaces. And in a particular field, the charge density describes how much the electric charge is accumulated.