Three concentric metallic shells $A$, $B$, and $C$ of radii $a$, $b$, and $c$ have surface charge densities $+\sigma$, $-\sigma$, and $\sigma+$ respectively. The potential of shell $B$ is
To find the potential of shell \(B\), we need to consider the potential due to the charge on each of the three shells \(A\), \(B\), and \(C\) at the location of shell \(B\)
Let \(V_A\), \(V_B\), and \(V_C\) be the potential due to shells \(A\), \(B\), and \(C\) respectively at shell \(B\). The total potential at shell \(B\) is then \(V = V_A + V_B + V_C\).
Potential due to shell A at shell B:
Since shell \(B\) is outside shell \(A\), the potential due to shell \(A\) at shell \(B\) is given by: \[V_A = \frac{Q_A}{4\pi\epsilon_0 b}\] where \(Q_A\) is the charge on shell \(A\). Since the surface charge density on shell \(A\) is \(\sigma\), the total charge \(Q_A\) is \(4\pi a^2 \sigma\). Therefore, \[V_A = \frac{4\pi a^2 \sigma}{4\pi\epsilon_0 b} = \frac{a^2 \sigma}{\epsilon_0 b}\]
Potential due to shell B at shell B:
The potential due to shell \(B\) at its own surface is: \[V_B = \frac{Q_B}{4\pi\epsilon_0 b}\] where \(Q_B\) is the charge on shell \(B\). Since the surface charge density on shell \(B\) is \(-\sigma\), the total charge \(Q_B\) is \(-4\pi b^2 \sigma\). Therefore, \[V_B = \frac{-4\pi b^2 \sigma}{4\pi\epsilon_0 b} = \frac{-b \sigma}{\epsilon_0}\]
Potential due to shell C at shell B:
Since shell \(B\) is inside shell \(C\), the potential due to shell \(C\) at shell \(B\) is the same as the potential at the surface of shell \(C\). Thus, \[V_C = \frac{Q_C}{4\pi\epsilon_0 c}\] where \(Q_C\) is the charge on shell \(C\). Since the surface charge density on shell \(C\) is \(\sigma\), the total charge \(Q_C\) is \(4\pi c^2 \sigma\). Therefore, \[V_C = \frac{4\pi c^2 \sigma}{4\pi\epsilon_0 c} = \frac{c \sigma}{\epsilon_0}\]
Total Potential at Shell B:
Adding the potentials due to each shell, we get: \[V = V_A + V_B + V_C = \frac{a^2 \sigma}{\epsilon_0 b} - \frac{b \sigma}{\epsilon_0} + \frac{c \sigma}{\epsilon_0} = \frac{\sigma}{\epsilon_0} \left( \frac{a^2}{b} - b + c \right)\]
Thus, the potential of shell \(B\) is \(\frac{\sigma}{\epsilon_0} \left( \frac{a^2}{b} - b + c \right)\), which corresponds to the answer \(\frac{(a^2/b - b + c) \sigma}{\epsilon_0}\).
