Question

The adjacent figure shows charged spherical shells A, B and C having charge densities $\sigma$, -$\sigma$, $\sigma$ and radii a, b, c respectively. If $V_A$=$V_C$ then c is equal to (in m) (assuming a = 0.10m, b = 0.20 m)

Answer 1

0.30

Answer 2

The potential at the surface of a charged spherical shell of radius $R$ and charge $Q$ is given by $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}$. The potential at a distance $r$ from the center due to a charged spherical shell of radius $R$ and charge $Q$ is $\frac{1}{4\pi\epsilon_0} \frac{Q}{r}$ for $r \ge R$ and $\frac{1}{4\pi\epsilon_0} \frac{Q}{R}$ for $r \le R$.

The charges on the shells A, B, and C are $Q_A = 4\pi a^2 \sigma$, $Q_B = 4\pi b^2 (-\sigma)$, and $Q_C = 4\pi c^2 \sigma$, respectively.

The potential at the surface of shell A ($r=a$) is the sum of the potentials due to shells A, B, and C at $r=a$.

$V_A = V_{A \text{ on } A} + V_{B \text{ on } A} + V_{C \text{ on } A}$

$V_{A \text{ on } A} = \frac{1}{4\pi\epsilon_0} \frac{Q_A}{a} = \frac{1}{4\pi\epsilon_0} \frac{4\pi a^2 \sigma}{a} = \frac{\sigma a}{\epsilon_0}$

Since $a < b$, the potential due to shell B at $r=a$ is the same as the potential on the surface of B.

$V_{B \text{ on } A} = \frac{1}{4\pi\epsilon_0} \frac{Q_B}{b} = \frac{1}{4\pi\epsilon_0} \frac{-4\pi b^2 \sigma}{b} = -\frac{\sigma b}{\epsilon_0}$

Since $a < c$, the potential due to shell C at $r=a$ is the same as the potential on the surface of C.

$V_{C \text{ on } A} = \frac{1}{4\pi\epsilon_0} \frac{Q_C}{c} = \frac{1}{4\pi\epsilon_0} \frac{4\pi c^2 \sigma}{c} = \frac{\sigma c}{\epsilon_0}$

So, $V_A = \frac{\sigma a}{\epsilon_0} - \frac{\sigma b}{\epsilon_0} + \frac{\sigma c}{\epsilon_0} = \frac{\sigma}{\epsilon_0} (a - b + c)$.

The potential at the surface of shell C ($r=c$) is the sum of the potentials due to shells A, B, and C at $r=c$.

$V_C = V_{A \text{ on } C} + V_{B \text{ on } C} + V_{C \text{ on } C}$

Since $c > a$, the potential due to shell A at $r=c$ is the same as the potential at a distance $c$ from the center.

$V_{A \text{ on } C} = \frac{1}{4\pi\epsilon_0} \frac{Q_A}{c} = \frac{1}{4\pi\epsilon_0} \frac{4\pi a^2 \sigma}{c} = \frac{\sigma a^2}{\epsilon_0 c}$

Since $c > b$, the potential due to shell B at $r=c$ is the same as the potential at a distance $c$ from the center.

$V_{B \text{ on } C} = \frac{1}{4\pi\epsilon_0} \frac{Q_B}{c} = \frac{1}{4\pi\epsilon_0} \frac{-4\pi b^2 \sigma}{c} = -\frac{\sigma b^2}{\epsilon_0 c}$

$V_{C \text{ on } C} = \frac{1}{4\pi\epsilon_0} \frac{Q_C}{c} = \frac{1}{4\pi\epsilon_0} \frac{4\pi c^2 \sigma}{c} = \frac{\sigma c}{\epsilon_0}$

So, $V_C = \frac{\sigma a^2}{\epsilon_0 c} - \frac{\sigma b^2}{\epsilon_0 c} + \frac{\sigma c}{\epsilon_0} = \frac{\sigma}{\epsilon_0} \left( \frac{a^2 - b^2}{c} + c \right)$.

Given $V_A = V_C$:

$\frac{\sigma}{\epsilon_0} (a - b + c) = \frac{\sigma}{\epsilon_0} \left( \frac{a^2 - b^2}{c} + c \right)$

Assuming $\sigma \neq 0$, we can cancel $\frac{\sigma}{\epsilon_0}$:

$a - b + c = \frac{a^2 - b^2}{c} + c$

$a - b = \frac{a^2 - b^2}{c}$

$c(a - b) = a^2 - b^2$

$c(a - b) = (a - b)(a + b)$

Since the radii are distinct ($a < b < c$), $a \neq b$, so $a - b \neq 0$. We can divide by $(a - b)$:

$c = a + b$

We are given $a = 0.10 \, \text{m}$ and $b = 0.20 \, \text{m}$.

$c = 0.10 \, \text{m} + 0.20 \, \text{m} = 0.30 \, \text{m}$.