Step 1: Analyze Forces and Accelerations

Consider two positive charges $q$ with masses $m$ and $M$ ($M > m$). Let the external electric field $E$ point in the positive x-direction.

Let $x_m$ and $x_M$ be the positions of the lighter and heavier masses respectively. The equations of motion are:

$$ m a_m = qE – \frac{kq^2}{r^2} $$ $$ M a_M = qE + \frac{kq^2}{r^2} $$

Here, $r = x_M – x_m$ is the separation, and $k = \frac{1}{4\pi\epsilon_0}$. Note that the mutual electrostatic force $F_{int} = \frac{kq^2}{r^2}$ pushes $m$ to the left (relative to $M$) and $M$ to the right.

Step 2: Analyze Relative Acceleration

We define the relative acceleration $a_{rel} = a_M – a_m$:

$$ a_{rel} = \left( \frac{qE}{M} + \frac{kq^2}{Mr^2} \right) – \left( \frac{qE}{m} – \frac{kq^2}{mr^2} \right) $$ $$ a_{rel} = qE \left( \frac{1}{M} – \frac{1}{m} \right) + \frac{kq^2}{r^2} \left( \frac{1}{M} + \frac{1}{m} \right) $$

Since $M > m$, the term $\left( \frac{1}{M} – \frac{1}{m} \right)$ is negative. Let us rewrite this using the reduced mass $\mu = \frac{mM}{M+m}$:

$$ a_{rel} = \frac{kq^2}{\mu r^2} – \frac{qE}{\mu} \left( \frac{M-m}{M+m} \right) $$

This equation describes the relative motion equivalent to a single particle of mass $\mu$ moving under a repulsive inverse-square force and a constant attractive force. Let the effective constant attractive force be $F_{eff} = qE \left( \frac{M-m}{M+m} \right)$.

Step 3: Determine the Equilibrium Separation

Before solving for the maximum separation, it is useful to identify the “equilibrium” separation $l_0$ where the relative acceleration is zero ($\ddot{r} = 0$).

$$ \frac{kq^2}{l_0^2} = F_{eff} = qE \left( \frac{M-m}{M+m} \right) $$ $$ l_0 = \sqrt{ \frac{kq(M+m)}{E(M-m)} } = \sqrt{ \frac{q(M+m)}{4\pi\epsilon_0 E(M-m)} } $$

Step 4: Calculate Maximum Separation using Energy Conservation

The particles are released from rest with an initial separation $l$. We consider two cases:

Case 1: $l > l_0$
If the initial separation is greater than the equilibrium separation, the effective attractive force (due to the field accelerating the lighter mass faster) dominates the electrostatic repulsion. The particles will move closer to each other initially. Since they start from rest, the maximum separation is the initial separation itself. $$ r_{max} = l $$

Case 2: $l < l_0$
If the initial separation is less than $l_0$, the electrostatic repulsion dominates. The particles will move apart. The relative velocity increases from zero and then decreases back to zero at the maximum separation $r_{max}$.

Using the Work-Energy theorem for relative motion (Change in KE = 0):

$$ \int_{l}^{r_{max}} F_{net, rel} \, dr = 0 $$ $$ \int_{l}^{r_{max}} \left( \frac{kq^2}{r^2} – F_{eff} \right) dr = 0 $$ $$ \left[ -\frac{kq^2}{r} – F_{eff} r \right]_{l}^{r_{max}} = 0 $$ $$ -\frac{kq^2}{r_{max}} – F_{eff} r_{max} – \left( -\frac{kq^2}{l} – F_{eff} l \right) = 0 $$ $$ kq^2 \left( \frac{1}{l} – \frac{1}{r_{max}} \right) = F_{eff} (r_{max} – l) $$ $$ kq^2 \frac{(r_{max} – l)}{l r_{max}} = F_{eff} (r_{max} – l) $$

Assuming $r_{max} \neq l$, we divide by $(r_{max} – l)$:

$$ \frac{kq^2}{l r_{max}} = F_{eff} $$ $$ r_{max} = \frac{kq^2}{l F_{eff}} $$

Substituting $F_{eff} = \frac{kq^2}{l_0^2}$ from Step 3:

$$ r_{max} = \frac{kq^2}{l (kq^2/l_0^2)} = \frac{l_0^2}{l} $$