Two charges

Consider two charges $q_1$ and $q_2$ of masses $m_1$ and $m_2$ respectively, which are free to move along the X-axis. $q_1$ is projected along the positive X axis from the origin with a speed $v_1$ while $q_2$ is projected along the negative X-axis with a speed of $v_2$ . Assume that $q_1$ and $q_2$ have the same sign.

Applying Newton's 2ns law for each charge at a general time $t$ leads to the equations:

$\ddot{x}_1 = -\frac{Kq_1q_2}{m_1(x_2-x_1)^2}$ $\ddot{x}_2 = \frac{Kq_1q_2}{m_2(x_2-x_1)^2}$

Subtracting both equations leads to:

$\ddot{x}_2 - \ddot{x}_1 =\frac{Kq_1q_2}{(x_2-x_1)^2}\left(\frac{1}{m_1} + \frac{1}{m_2}\right)$

Let: $x_2 - x_1 = x$ $\implies \ddot{x}_2 - \ddot{x}_1=\ddot{x}$ and:

$P=Kq_1q_2\left(\frac{1}{m_1} + \frac{1}{m_2}\right)$

This simplifies the equation of motion to:

$\ddot{x} = \frac{P}{x^2}$ $\dot{x} \frac{d \dot{x}}{dx} = \frac{P}{x^2}$ $\frac{\dot{x}^2}{2} = -\frac{P}{x} + C$

Assuming that the initial separation is $x(0) = d$ and $\dot{x}_2(0)-\dot{x}_1(0) =\dot{x}(0) -v_2-v_1$ . These initial conditions allow us to solve for $C$ .

Using the initial conditions allows us to arrive at a unique solution for $x$ as a function of time. I assume that the initial conditions of the system are such that the charges do not collide at any point of time. It can be shown by continuing the above analysis that $x$ is continuous function of time. For $x$ to be minimum at some time $t$ , it must hold true at that $\dot{x}$ is equal to zero at that instant. Thus when $x = x_{min}$ :

$\implies \dot{x}_2-\dot{x}_1=0$ $\implies \dot{x}_2=\dot{x}_1$

This condition also allows us to derive the expression for the minimum separation as such:

$0= -\frac{P}{x_{min}} + C$