© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the charges on each shell
• Shell A (radius = $a$) has surface charge density $+\sigma$.
• Shell B (radius = $b$) has surface charge density $-\sigma$.
• Shell C (radius = $c$) has surface charge density $+\sigma$.
Here, $a < b < c$.
Step 2: Express the total charge on each shell
The total charge $Q$ on a spherical shell of radius $R$ having surface charge density $\sigma$ is given by
$$Q = 4\pi R^2 \sigma.$$
Thus,
• Charge on shell A: $Q_A = 4\pi a^2 (+\sigma) = 4\pi a^2 \sigma.$
• Charge on shell B: $Q_B = 4\pi b^2 (-\sigma) = -4\pi b^2 \sigma.$
• Charge on shell C: $Q_C = 4\pi c^2 (+\sigma) = 4\pi c^2 \sigma.$
Step 3: Recall the potential due to a spherical shell
1. Outside (or on the surface of) a charged spherical shell of radius $R$, the potential at a distance $r \ge R$ is
$$V = \frac{1}{4\pi \epsilon_0}\,\frac{Q}{r}.$$
2. Inside the shell ($r < R$), the potential is constant and equals the value at $r = R$, which is
$$V = \frac{1}{4\pi \epsilon_0}\,\frac{Q}{R}.$$
Step 4: Compute potential at shell B due to each shell
Since shell B has radius $b$, let us calculate the potential at $r = b$ due to all charges:
Potential at B due to shell A
Since $b > a$, shell B is outside shell A. Thus,
$$V_{BA} = \frac{1}{4\pi\epsilon_0}\,\frac{Q_A}{b}
= \frac{1}{4\pi\epsilon_0} \cdot \frac{4\pi a^2 \sigma}{b}
= \frac{\sigma a^2}{\epsilon_0\,b}.$$
Potential at B due to shell B itself
The potential at the surface of a charged spherical shell of radius $b$ is
$$V_{BB} = \frac{1}{4\pi\epsilon_0}\,\frac{Q_B}{b}
= \frac{1}{4\pi\epsilon_0} \cdot \frac{-4\pi b^2 \sigma}{b}
= -\,\frac{\sigma b}{\epsilon_0}.$$
Potential at B due to shell C
Since $b < c$, shell B is inside shell C. Inside a charged shell, the potential is constant and equals the surface potential of shell C:
$$V_{BC} = \frac{1}{4\pi\epsilon_0}\,\frac{Q_C}{c}
= \frac{1}{4\pi\epsilon_0} \cdot \frac{4\pi c^2 \sigma}{c}
= \frac{\sigma c}{\epsilon_0}.$$
Step 5: Sum up all contributions
The total potential on shell B is the sum of the above three contributions:
$$
V_B
= V_{BA} + V_{BB} + V_{BC}
= \frac{\sigma a^2}{\epsilon_0\,b} - \frac{\sigma b}{\epsilon_0} + \frac{\sigma c}{\epsilon_0}.
$$
Factor out $\frac{\sigma}{\epsilon_0}$:
$$
V_B
= \frac{\sigma}{\epsilon_0}\left(\frac{a^2}{b} - b + c\right)
= \frac{\sigma}{\epsilon_0}\left[\frac{a^2 - b^2}{b} + c\right].
$$
Step 6: Final answer
Hence, the potential of shell B is:
$$
\boxed{\frac{\sigma}{\epsilon_0}\left[\frac{a^2 - b^2}{b} + c\right]}.
$$
