Question

The figure shows a conducting sphere 'A' of radius 'a' which is surrounded by a neutral conducting spherical shell B of radius 'b' (> a). Initially switches $S_1, S_2$ and $S_3$ are open and sphere 'A' carries a charge $Q$. First the switch '$S_1$' is closed to connect the shell B with the ground and then opened. Now the switch '$S_2$' is closed so that the sphere 'A' is grounded and then $S_2$ is opened. Finally, the switch '$S_3$' is closed to connect the spheres together. If the heat (in mJ) which is produced after closing the switch $S_3$ is $x$, then find $\frac{x}{9}$. [Consider $b = 4$ cm, $a = 2$ cm and $Q = 0.8\mu C$]

Answer 1

2

Answer 2

The problem involves a sequence of electrostatic processes. Let's analyze the charge distribution and potential energy at each stage.

Initial state: Sphere A has charge $Q$, shell B is neutral. Switches $S_1, S_2, S_3$ are open.

Step 1: $S_1$ is closed to ground shell B, then opened. When B is grounded, its potential becomes zero. Sphere A has charge Q. Charge $-Q$ is induced on the inner surface of B. When grounded, the outer surface charge goes to the earth. So, after grounding B, the shell B has a total charge of $-Q$, distributed on its inner surface. Sphere A still has charge Q.

Step 2: $S_2$ is closed to ground sphere A, then opened. Now sphere A is grounded, so its potential is zero. Shell B has charge $-Q$ on its inner surface. Let the new charge on A be $Q'$. The potential of A is $V_A = \frac{1}{4\pi\epsilon_0} \frac{Q'}{a} + \frac{1}{4\pi\epsilon_0} \frac{-Q}{b}$. Since $V_A = 0$, we have $\frac{Q'}{a} = \frac{Q}{b}$, so $Q' = Q \frac{a}{b}$. After this step, sphere A has charge $Q_A = Q \frac{a}{b}$, and shell B has charge $Q_B = -Q$ on its inner surface. The outer surface of B has zero charge.

Step 3: $S_3$ is closed to connect A and B. When A and B are connected, they form a single conductor. Charge will redistribute until they are at the same potential. Since A is inside B and they are connected, the combined charge will reside on the outer surface of the system. The total charge on the combined system is $Q_{total} = Q_A + Q_B = Q \frac{a}{b} + (-Q) = Q \left( \frac{a}{b} - 1 \right)$. This charge will reside on the outer surface of shell B. The final configuration is a spherical shell of radius 'b' with charge $Q_{final} = Q \left( \frac{a}{b} - 1 \right)$ on its outer surface.

The heat produced when $S_3$ is closed is the loss in potential energy: $x = U_{before\ S_3} - U_{after\ S_3}$.

Before closing $S_3$, sphere A has charge $Q' = Q \frac{a}{b}$ and shell B has charge $-Q$ on its inner surface. The potential energy before $S_3$ is the energy stored in the electric field between A and B. For $a < r < b$, the electric field is $E(r) = \frac{Q'}{4\pi\epsilon_0 r^2}$. The potential energy in this region is $U_{ab} = \int_a^b \frac{1}{2} \epsilon_0 E^2 dV = \int_a^b \frac{1}{2} \epsilon_0 \left( \frac{Q'}{4\pi\epsilon_0 r^2} \right)^2 4\pi r^2 dr = \frac{(Q')^2}{8\pi\epsilon_0} \int_a^b \frac{1}{r^2} dr = \frac{(Q')^2}{8\pi\epsilon_0} \left( \frac{1}{a} - \frac{1}{b} \right)$. Substituting $Q' = Q \frac{a}{b}$, $U_{before\ S_3} = \frac{(Q a/b)^2}{8\pi\epsilon_0} \frac{b-a}{ab} = \frac{Q^2 a^2/b^2}{8\pi\epsilon_0} \frac{b-a}{ab} = \frac{Q^2 a (b-a)}{8\pi\epsilon_0 b^3}$.

The potential energy after closing $S_3$ is $U_{after\ S_3} = \frac{1}{2} \frac{\left( Q \left( \frac{a}{b} - 1 \right) \right)^2}{4\pi\epsilon_0 b} = \frac{Q^2 (a/b - 1)^2}{8\pi\epsilon_0 b} = \frac{Q^2 (a-b)^2/b^2}{8\pi\epsilon_0 b} = \frac{Q^2 (b-a)^2}{8\pi\epsilon_0 b^3}$.

The heat produced is $x = U_{before\ S_3} - U_{after\ S_3} = \frac{Q^2 a (b-a)}{8\pi\epsilon_0 b^3} - \frac{Q^2 (b-a)^2}{8\pi\epsilon_0 b^3} = \frac{Q^2 (b-a)}{8\pi\epsilon_0 b^3} [a - (b-a)] = \frac{Q^2 (b-a)}{8\pi\epsilon_0 b^3} (2a - b)$.

Given values: $a = 2$ cm = $0.02$ m, $b = 4$ cm = $0.04$ m, $Q = 0.8 \mu C = 0.8 \times 10^{-6} C$. $b-a = 0.04 - 0.02 = 0.02$ m. $2a-b = 2(0.02) - 0.04 = 0.04 - 0.04 = 0$.

So, $x = \frac{Q^2 a(b-a)}{8\pi\epsilon_0 b^3}$. $x = \frac{(0.8 \times 10^{-6})^2 \times 0.02 \times 0.02}{8 \times \frac{1}{9 \times 10^9} \times (0.04)^3} = \frac{0.64 \times 10^{-12} \times 0.0004 \times 9 \times 10^9}{8 \times 0.000064}$. $x = \frac{0.64 \times 4 \times 10^{-16} \times 9 \times 10^9}{8 \times 64 \times 10^{-6}} = \frac{2.56 \times 9 \times 10^{-7}}{512 \times 10^{-6}} = \frac{23.04 \times 10^{-1}}{512} = \frac{2.304}{512} = 0.0045$. $x = 0.0045 J = 4.5 mJ$.

Where is the error?

Let's check if the formula is correct: $x = \frac{Q^2 a(b-a)}{8\pi\epsilon_0 b^3}$. $8\pi\epsilon_0 = \frac{2}{9 \times 10^9}$. $x = \frac{(0.8 \times 10^{-6})^2 \times 0.02 \times 0.02}{\frac{2}{9 \times 10^9} \times (0.04)^3} = \frac{0.64 \times 10^{-12} \times 0.0004 \times 9 \times 10^9}{2 \times 0.000064} = \frac{0.64 \times 4 \times 10^{-16} \times 9 \times 10^9}{2 \times 64 \times 10^{-6}}$. $x = \frac{2.56 \times 9 \times 10^{-7}}{128 \times 10^{-6}} = \frac{23.04 \times 10^{-1}}{128} = \frac{2.304}{128} = 0.018$. $x = 0.018 J = 18 mJ$.

So $x = 18$ mJ. $\frac{x}{9} = \frac{18}{9} = 2$.