Consider two particles with charges $+q_1$ and $-q_2$ and masses $m_1$ and $m_2$ respectively, placed in a uniform external electric field $\vec{E}$.

Let $\vec{E}$ be directed along the positive x-axis. The forces acting on the particles are:

• Particle 1: Experiences an electric force $q_1E$ (forward) and a Coulomb attraction force $F_c$ (forward, assuming Particle 2 leads, or balancing the drag).
• Particle 2: Experiences an electric force $-q_2E$ (backward) and a Coulomb attraction force $-F_c$ (backward).

Using Newton’s Second Law ($\vec{F}_{net} = m\vec{a}$) for both particles, assuming a common acceleration $a$:

$$ m_1 a = q_1 E + F_{Coulomb} \quad \dots (1) $$ $$ m_2 a = -q_2 E – F_{Coulomb} \quad \dots (2) $$

Note: We express the equations vectorially. Since the charges are opposite, the external field tends to pull them apart (accelerating $+q_1$ right and $-q_2$ left). The internal Coulomb force is attractive, pulling them together. For the separation to remain constant, the net acceleration of both particles must be identical (both moving in one direction, or both stationary).

To eliminate acceleration $a$, we can equate the acceleration expressions:

$$ \frac{q_1 E + F_c}{m_1} = \frac{-q_2 E – F_c}{m_2} $$

Rearranging to solve for the Coulomb force magnitude $F_c$:

$$ \frac{F_c}{m_1} + \frac{F_c}{m_2} = -\frac{q_2 E}{m_2} – \frac{q_1 E}{m_1} $$

Taking the magnitude (since $F_c$ is attractive and opposes the separation tendency):

$$ F_c \left( \frac{1}{m_1} + \frac{1}{m_2} \right) = E \left( \frac{q_1}{m_1} + \frac{q_2}{m_2} \right) $$ $$ F_c \left( \frac{m_1 + m_2}{m_1 m_2} \right) = E \left( \frac{q_1 m_2 + q_2 m_1}{m_1 m_2} \right) $$

Canceling the $m_1 m_2$ term:

$$ F_c (m_1 + m_2) = E (q_1 m_2 + q_2 m_1) $$

Substitute Coulomb’s Law $F_c = \frac{k q_1 q_2}{r^2}$, where $k = \frac{1}{4\pi\epsilon_0}$:

$$ \frac{k q_1 q_2}{r^2} (m_1 + m_2) = E (q_1 m_2 + q_2 m_1) $$

Solving for the separation distance $r$:

$$ r^2 = \frac{k q_1 q_2 (m_1 + m_2)}{E (q_1 m_2 + q_2 m_1)} $$