Physics Principles Involved:

• Joule Heating: The heat energy dissipated in a resistor due to current flow.
• Energy Conservation: Analyzing the energy transfer associated with charge flow at constant potential.

From the previous analysis (Question 20), we established that the potential of the balloon is maintained at a constant value \( V_0 \). The balloon is connected to the ground via a resistor \( R \). According to Ohm’s Law, the current \( I \) flowing through the resistor is constant: $$ I = \frac{V_0}{R} $$ The rate at which heat is dissipated in the resistor (Power, \( P \)) is given by Joule’s Law: $$ P = I^2 R = \left( \frac{V_0}{R} \right)^2 R = \frac{V_0^2}{R} $$ Since \( V_0 \) and \( R \) are constants, the power dissipation is constant throughout the process.

We need to find the total heat dissipated until the radius of the balloon becomes half. Let the initial radius be \( a \) and the final radius be \( a/2 \).
From the solution to Question 20, we know the magnitude of the rate of change of radius is constant: $$ v = \left| \frac{dr}{dt} \right| = \frac{1}{4\pi\varepsilon_0 R} $$ The total change in radius (distance “traveled” by the shrinking surface) is: $$ \Delta r = a – \frac{a}{2} = \frac{a}{2} $$ Since the rate is constant, the time interval \( \Delta t \) required for this change is simply distance divided by speed: $$ \Delta t = \frac{\Delta r}{v} = \frac{a/2}{1 / (4\pi\varepsilon_0 R)} $$ $$ \Delta t = \frac{a}{2} \cdot (4\pi\varepsilon_0 R) = 2\pi\varepsilon_0 a R $$

The total heat \( H \) dissipated is the product of the constant power and the time duration: $$ H = P \times \Delta t $$ Substituting the expressions derived in Step 1 and Step 2: $$ H = \left( \frac{V_0^2}{R} \right) \times (2\pi\varepsilon_0 a R) $$ $$ H = 2\pi\varepsilon_0 a V_0^2 $$

Heat dissipated can also be viewed as the loss of potential energy of the charge that flows from potential \( V_0 \) to ground (potential 0). $$ H = \int V dq = V_0 \int dq = V_0 \Delta Q $$ The initial charge is \( Q_i = 4\pi\varepsilon_0 a V_0 \).
The final charge (when radius is \( a/2 \)) is \( Q_f = 4\pi\varepsilon_0 (a/2) V_0 \).
The total charge \( \Delta Q \) that flows through the resistor is: $$ \Delta Q = Q_i – Q_f = 4\pi\varepsilon_0 V_0 (a – a/2) = 2\pi\varepsilon_0 a V_0 $$ Therefore, the heat dissipated is: $$ H = V_0 (2\pi\varepsilon_0 a V_0) = 2\pi\varepsilon_0 a V_0^2 $$