In the steady state, the spheres connected to the circuit are fully charged, and no current flows into or out of them. A steady current $I$ flows through the series combination of $n$ resistors. The spheres are isolated from each other and are essentially at the potential of the node they are connected to.

The total resistance of the circuit is $R_{eq} = nR$.

The current in the circuit is: $$ I = \frac{V_0}{nR} $$

The rightmost terminal is grounded, so its potential is $0$. The leftmost terminal is connected to the positive terminal of the battery, so its potential is $V_0$.

There are $n$ resistors forming a potential divider. The potential drops linearly from left to right. Let’s index the nodes (and spheres) from $k=0$ to $k=n$.

• At $k=0$ (leftmost): Potential $V_0 = V_0$
• At $k=n$ (rightmost): Potential $V_n = 0$

The potential at the $k$-th node (after $k$ resistors) is given by:

$$ V_k = V_0 – k(IR) = V_0 – k\left(\frac{V_0}{nR}\right)R = V_0 \left( 1 – \frac{k}{n} \right) $$

The capacitance of an isolated sphere of radius $r$ is $C = 4\pi\varepsilon_0 r$. The charge on the $k$-th sphere is:

$$ q_k = C V_k = 4\pi\varepsilon_0 r V_0 \left( 1 – \frac{k}{n} \right) $$

The total charge $Q$ is the sum of charges on all $n+1$ spheres (from $k=0$ to $n$):

$$ Q = \sum_{k=0}^{n} q_k = 4\pi\varepsilon_0 r V_0 \sum_{k=0}^{n} \left( 1 – \frac{k}{n} \right) $$

Let’s evaluate the sum:

$$ \sum_{k=0}^{n} \left( 1 – \frac{k}{n} \right) = \sum_{k=0}^{n} 1 – \frac{1}{n}\sum_{k=0}^{n} k $$

The first term sums $1$, $(n+1)$ times. The second term is the sum of the first $n$ integers:

$$ = (n+1) – \frac{1}{n} \frac{n(n+1)}{2} = (n+1) – \frac{n+1}{2} = \frac{n+1}{2} $$

Substituting this back into the charge equation:

$$ Q = 4\pi\varepsilon_0 r V_0 \left[ \frac{n+1}{2} \right] $$ $$ Q = 2\pi\varepsilon_0 r (n+1) V_0 $$

The total charge accumulated on the spheres is:

$$ \mathbf{2\pi\varepsilon_0 r (n+1) V_0} $$