Physics Principles Involved:

• Electrostatics of Conductors: The relationship between charge, potential, and radius for an isolated spherical conductor.
• Current Electricity: Ohm’s Law and the definition of electric current as the rate of flow of charge.
• Differentiation: Relating rates of change of physical quantities.

We are given an isolated spherical conductor (the balloon) of instantaneous radius \( r \) carrying a charge \( q \). The electric potential \( V \) of a sphere is given by the formula: $$ V = \frac{q}{4\pi\varepsilon_0 r} $$ The problem states that the balloon is controlled such that its potential remains constant at \( V_0 \) throughout the process. Therefore, we can express the instantaneous charge \( q(t) \) in terms of the constant potential and the time-varying radius \( r(t) \): $$ q(t) = 4\pi\varepsilon_0 V_0 r(t) $$

As the balloon shrinks, charge must leave the surface to maintain the constant potential \( V_0 \). By differentiating the charge equation with respect to time \( t \), we get: $$ \frac{dq}{dt} = 4\pi\varepsilon_0 V_0 \frac{dr}{dt} $$ Since the radius is decreasing, \( \frac{dr}{dt} \) is negative, and consequently, \( \frac{dq}{dt} \) is negative, indicating a loss of charge.

The balloon is connected to the ground through a resistor \( R \). Since the balloon is at potential \( V_0 \) and the ground is at zero potential, an electric current \( I \) flows from the balloon to the ground. According to Ohm’s Law: $$ I = \frac{V_0 – 0}{R} = \frac{V_0}{R} $$ Electric current is defined as the rate of flow of charge. Since charge is leaving the balloon, the current is equal to the negative rate of change of charge: $$ I = -\frac{dq}{dt} $$ Substituting the expression for \( \frac{dq}{dt} \) from Step 2: $$ I = – \left( 4\pi\varepsilon_0 V_0 \frac{dr}{dt} \right) $$

Now we equate the two expressions for current \( I \): $$ \frac{V_0}{R} = -4\pi\varepsilon_0 V_0 \frac{dr}{dt} $$ Notice that the potential \( V_0 \) cancels out from both sides. We can now rearrange the equation to solve for the rate of change of the radius, \( \frac{dr}{dt} \): $$ \frac{dr}{dt} = -\frac{1}{4\pi\varepsilon_0 R} $$ The magnitude of the rate of change (the speed at which the radius decreases) is: $$ \left| \frac{dr}{dt} \right| = \frac{1}{4\pi\varepsilon_0 R} $$ This result indicates that the radius decreases at a constant rate, independent of the instantaneous radius \( r \) or the initial radius \( a \).