When switch $S_1$ is closed, the inner shell (radius $r$) and the outer shell (radius $3r$) are connected by a conducting wire. This implies they form a single isolated conductor. Therefore, their electric potentials must be equal: $$ V_{inner} = V_{outer} $$ Let the charge on the inner shell be $q$. Since the system is initially neutral and isolated, the total charge must remain zero. Thus, the charge on the outer shell is $-q$.

The potential at the surface of the inner shell ($V_{in}$) is the sum of potentials due to its own charge $q$, the outer shell’s charge $-q$, and the external point charge $+Q$ located at distance $2r$: $$ V_{in} = \frac{kq}{r} + \frac{k(-q)}{3r} + \frac{kQ}{2r} $$ The potential at the surface of the outer shell ($V_{out}$) is due to the inner charge $q$ (acting as if at the center), the outer charge $-q$, and the point charge $+Q$ (which is effectively inside the outer shell’s sphere). Since the outer shell is a conductor enclosing $Q$, and there are no charges outside it, the potential is constant and determined by the net enclosed charge and geometry. However, a simpler approach is to calculate the potential at the surface due to all charges. For a point on the outer surface (radius $3r$), $Q$ acts as an internal charge. The potential due to $Q$ on the surface $3r$ is $kQ/3r$: $$ V_{out} = \frac{kq}{3r} + \frac{k(-q)}{3r} + \frac{kQ}{3r} $$ Equating $V_{in} = V_{out}$: $$ \frac{kq}{r} – \frac{kq}{3r} + \frac{kQ}{2r} = \frac{kq}{3r} – \frac{kq}{3r} + \frac{kQ}{3r} $$ $$ kq \left( \frac{1}{r} – \frac{1}{3r} \right) + \frac{kQ}{2r} = \frac{kQ}{3r} $$ $$ q \left( \frac{2}{3r} \right) = Q \left( \frac{1}{3r} – \frac{1}{2r} \right) $$ $$ q \left( \frac{2}{3r} \right) = Q \left( \frac{2-3}{6r} \right) = -\frac{Q}{6r} $$ $$ q = -\frac{Q}{6r} \cdot \frac{3r}{2} = -\frac{Q}{4} $$ Thus, the inner shell acquires charge $q_{in} = -Q/4$ and the outer shell acquires $q_{out} = +Q/4$.

Heat dissipated is equal to the loss in the total electrostatic potential energy of the system ($H = U_i – U_f$).

Initial State ($U_i$): Both shells are neutral ($q=0$). The only energy is the self-energy of $Q$ (which is constant) and interaction terms. Since shells have 0 charge, interaction energy is zero. $$ U_i = 0 $$ (Note: We ignore the infinite self-energy of the point charge as it cancels out).

Final State ($U_f$): The common potential of the shells is: $$ V_{common} = V_{out} = \frac{kQ}{3r} $$ The interaction energy of the system is given by: $$ U_f = \frac{1}{2}q_{in}V_{in} + \frac{1}{2}q_{out}V_{out} + \frac{1}{2}Q V’_{at\_Q} $$ Since $V_{in}=V_{out}$ and $q_{in} + q_{out} = 0$, the first two terms sum to zero. We only calculate the interaction energy of $Q$ with the induced charges. The potential at position $2r$ ($V’_{at\_Q}$) due to the shells is: $$ V’_{at\_Q} = \frac{k(q_{in})}{2r} + \frac{k(q_{out})}{3r} $$ $$ V’_{at\_Q} = \frac{k(-Q/4)}{2r} + \frac{k(Q/4)}{3r} = -\frac{kQ}{8r} + \frac{kQ}{12r} = kQ\left(\frac{2-3}{24r}\right) = -\frac{kQ}{24r} $$ Thus, $$ U_f = \frac{1}{2} Q \left( -\frac{kQ}{24r} \right) = -\frac{kQ^2}{48r} $$

$$ H = U_i – U_f = 0 – \left( -\frac{kQ^2}{48r} \right) = \frac{kQ^2}{48r} $$ Substituting $k = \frac{1}{4\pi\epsilon_0}$: $$ H = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{48r} = \frac{Q^2}{192\pi\epsilon_0 r} $$

Closing switch $S_2$ grounds the outer shell, so $V_{out} = 0$. Since $S_1$ remains closed, the inner shell is connected to the outer shell, implying $V_{in} = V_{out} = 0$.

Let the new charges be $q_1$ (inner) and $q_2$ (outer). We apply the potential equations for $V_{in} = 0$ and $V_{out} = 0$.

Equation 1 (Outer Surface Potential): $$ V_{out} = \frac{k(q_1 + q_2)}{3r} + \frac{kQ}{3r} = 0 $$ $$ \implies q_1 + q_2 = -Q $$ Equation 2 (Inner Surface Potential): $$ V_{in} = \frac{kq_1}{r} + \frac{kq_2}{3r} + \frac{kQ}{2r} = 0 $$ Substituting $q_2 = -Q – q_1$ into Equation 2: $$ \frac{kq_1}{r} + \frac{k(-Q – q_1)}{3r} + \frac{kQ}{2r} = 0 $$ $$ \frac{q_1}{r} – \frac{Q}{3r} – \frac{q_1}{3r} + \frac{Q}{2r} = 0 $$ $$ q_1 \left( \frac{1}{r} – \frac{1}{3r} \right) = Q \left( \frac{1}{3r} – \frac{1}{2r} \right) $$ $$ q_1 \left( \frac{2}{3r} \right) = Q \left( -\frac{1}{6r} \right) \implies q_1 = -\frac{Q}{4} $$

Compare the charges before and after closing $S_2$:

• Inner Shell Charge: Before: $-Q/4$. After: $-Q/4$.
  Change is zero. No current flows through $R_1$. Heat in $R_1$ is 0.
• Outer Shell Charge: Before: $+Q/4$. After: $q_2 = -Q – (-Q/4) = -3Q/4$.
  Charge flows through $R_2$ to ground. Heat is dissipated in $R_2$.

Initial Energy ($U_i$): From the end of Question 24, we found: $$ U_i = -\frac{kQ^2}{48r} $$ Final Energy ($U_f$): With potentials $V_{in}=V_{out}=0$, the self-energy terms of the shells vanish. The total energy is just the interaction energy of $Q$ with the induced charges ($q_1 = -Q/4$ at $r$ and $q_2 = -3Q/4$ at $3r$). $$ V’_{at\_Q} = \frac{kq_1}{2r} + \frac{kq_2}{3r} = \frac{k(-Q/4)}{2r} + \frac{k(-3Q/4)}{3r} $$ $$ V’_{at\_Q} = -\frac{kQ}{8r} – \frac{kQ}{4r} = -\frac{3kQ}{8r} $$ $$ U_f = \frac{1}{2} Q V’_{at\_Q} = \frac{1}{2} Q \left( -\frac{3kQ}{8r} \right) = -\frac{3kQ^2}{16r} $$

$$ H = U_i – U_f = \left( -\frac{kQ^2}{48r} \right) – \left( -\frac{3kQ^2}{16r} \right) $$ $$ H = \frac{kQ^2}{r} \left( \frac{3}{16} – \frac{1}{48} \right) = \frac{kQ^2}{r} \left( \frac{9-1}{48} \right) = \frac{kQ^2}{r} \left( \frac{8}{48} \right) = \frac{kQ^2}{6r} $$ Substituting $k = \frac{1}{4\pi\epsilon_0}$: $$ H = \frac{Q^2}{6r(4\pi\epsilon_0)} = \frac{Q^2}{24\pi\epsilon_0 r} $$

Since no heat is dissipated in $R_1$, this entire amount is dissipated in $R_2$.