Speed of a Point Charge Repelled from Another

Problem Statement

A positive charge (Q = +5.0 $\mu C$ and m = 0.5 kg) is fixed in place at the origin. Another positive charge (q = +1.0 $\mu C$ and m = 2.0 kg) is free to move and starts at rest at the point (1.0 m, 0.0 m). What is the speed of q when it is very far away from Q?

Solution

We can use the law of Conservation of Energy to solve this problem. Originally, the two point charges are separated by a distance of 1.0 m and are at rest. Thus, the total energy of the system is the electric potential energy of the two point charges:

$$\begin{array}{rl}U& =\frac{kQq}{r}\\ & =\frac{(8.99\times 10^{9}N{m}^2/C^2)(+5.0\mu C)(+1.0\mu C)}{1.0m}\\ & =0.04495J\end{array}$$

When the point charges are “very far apart”, the electric potential energy between them is equal to zero since the energy is inversely proportional to the distance between the charges. Thus, all of the initial potential energy gets converted into kinetic energy. Since only q is free to move:

$$0.04495J=K{E}_q=\frac{1}{2}(2.0kg)v^2$$

Thus, we get that:

$$v=0.21m/s$$