1. Circuit Analysis

In the steady state, the spheres do not draw any current. Therefore, a steady current $I$ flows through the resistor loop connected to the battery $V_0$.

The circuit consists of two equal resistors $R$ in series connected to the battery. The potential difference across the entire chain is $V_0$.

Let’s define the potentials at the terminals A, B, and C:

• The total voltage drop is $V_0$ across $2R$.
• $V_A – V_C = V_0$.
• Point B is exactly in the middle of the resistors. By symmetry, its potential is the average of A and C.

If we set the potential of the middle point B to be $V$, then:

$$V_A = V + \frac{V_0}{2}$$

$$V_C = V – \frac{V_0}{2}$$

2. Applying Charge Conservation

The problem states the spheres are initially uncharged and the “total charge on the spheres… gets redistributed”. This implies the system of three spheres is electrically isolated as a whole, so the net charge must remain zero:

Using the capacitance formula for isolated spheres ($Q = 4\pi\epsilon_0 r V_{sphere}$), we can write:

$$4\pi\epsilon_0 a V_A + 4\pi\epsilon_0 b V_B + 4\pi\epsilon_0 a V_C = 0$$

Substituting the potentials ($V_B = V$):

$$a(V + \frac{V_0}{2}) + b(V) + a(V – \frac{V_0}{2}) = 0$$

$$aV + a\frac{V_0}{2} + bV + aV – a\frac{V_0}{2} = 0$$

$$(2a + b)V = 0 \implies V = 0$$

This result implies that the potential of the middle sphere B settles at 0 (relative to infinity).

3. Calculating Final Charges

Now that we know $V_B = 0$, the potentials are:

• $V_B = 0$
• $V_A = +V_0/2$
• $V_C = -V_0/2$

Calculating the charges:

$$Q_A = C_A V_A = (4\pi\epsilon_0 a) \left(\frac{V_0}{2}\right) = 2\pi\epsilon_0 a V_0$$

$$Q_B = C_B V_B = (4\pi\epsilon_0 b) (0) = 0$$

$$Q_C = C_C V_C = (4\pi\epsilon_0 a) \left(-\frac{V_0}{2}\right) = -2\pi\epsilon_0 a V_0$$