The electric potential energy of a pair of point charges \(q_{l}\) \(q_{2}\) and at a distance \(r\) from each other can be expressed as \(U=\frac{k q_{1} q_{2}}{r_{12}}\)
If the charges are of different signs as in the case at hand, then \(U\) is negative and negative work is done by an external agent against the attractive force between the charges of opposite signs as they are brought near each other; a force must be applied opposite the displacement to prevent \(q_{1}\) (the electron) from accelerating toward \(q_{2}\) (the proton).
If the system consists of more than two charged particles, we can obtain the total potential energy of the system by calculation \(U\) for every pair of charges and summing the terms algebraically.
i.e. \(U=k\left(\frac{q_{1} q_{2}}{r_{12}}+\frac{q_{1} q_{3}}{r_{13}}+\frac{q_{2} q_{3}}{r_{23}}+\ldots\right)\)
\begin{aligned}
&U=\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} d} \\
&=\frac{1.6 \times 10^{-19}}{4 \times 3.142 \times 8.85 \times 10^{-12}} \times \frac{1.6 \times 10^{-19}}{0.5 \times 10^{-10}}
\end{aligned}
\begin{aligned}
&=\frac{1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{5.56134 \times 10^{-2 b}} \\
&=\frac{2.56 \times 10^{-38}}{5.56134 \times 10^{-21}}=4.6 \times 10^{-18} J
\end{aligned}