[quote]Imagine two charged particles are placed a small distance, say 1 mm, apart and held there. The charge on each of the particles  is 0.02C.

One of the particles is released. Describe the motion of the free particle and calculate the total work done.[/quote]

so;
$q_1 = q_2 = 0.02 \mathrm{C}$
$r = 1\times10^{-3} \mathrm{m}$
$F(r) = \frac{k \cdot q_1 q_2}{r^2}\mathrm{N} \text{ where } k = \frac{1}{4\pi\epsilon_0}$ $= 8.9875517873681764\times10^9 \mathrm{N \cdot m^2\cdot C^{-2}}$

We know that as $r$ increases $F$ decreases so,
$F(r)$ > $F(r+\delta r)$

We shall assume it is $q_2$ that is released and $q_1$ is held by an imaginary force.
When $q_2$ is released a force of $F(r) = \frac{k \cdot 0.02 \times 0.02}{(1\times10^{-3})^2}\mathrm{N}$ is pushing it in the positive direction.

After a tiny bit of time, $\delta t$, the particle will have moved $\delta r$ and so the force will have decreased meaning that the acceleration of the particle will have decreased...

This is where I get stuck.
how can we express the particles motion if it has no set mass...?

I was thinking change in momentum, but that still depends on mass...

This is an interesting one!
Please can you give me a pointer or two?

Is it actually possible to work it out without mass???