Electrostatic Energy from Discrete Charges

Section 16.2 Electrostatic Energy from Discrete Charges

How much energy is stored in a collection of discrete charges? The energy stored is equal to the work it takes to assemble the collection of charges. Work is the line integral of force times distance, which in this context takes the form

\begin{gather*} W = \int \FF\cdot d\rr = -q \int \EE\cdot d\rr = q \> \> \Delta V . \end{gather*}

Activity 16.2.1. Finding the electrostatic energy due to a collection of charges.

Imagine an empty room. One at a time, your friends enter this room, and take their seats. Pretend that each friend represents a charge, being brought in from infinity. As each charge is moved into its position, compute the energy (work) needed to bring this charge from its starting point to its final location. Write down an expression for the total energy when all charges have been moved into position.

Answer.

\begin{equation} U = \frac12 \frac{1}{4\pi\epsilon_0} \sum_{i\ne j} \frac{q_i q_j}{|\rr_j-\rr_i|}\tag{16.2.5} \end{equation}

Solution.

As you work through the computation of the total energy in the final charge configuration, you will quickly appreciate the need for a compact notation, in this case the summation notation.

What is the potential when no charges are present? \(V=0\text{,}\) so no energy is needed to bring the first charge in.

What is the potential when the first charge, \(q_1\text{,}\) is brought in to position \(\rr_1\text{?}\) We have

\begin{equation} V_1 = \frac{1}{4\pi\epsilon_0} \frac{q_1}{|\rr-\rr_1|}\tag{16.2.1} \end{equation}

so the energy needed to bring in the second charge, \(q_2\text{,}\) to position \(\rr_2\) is

\begin{equation} q_2 V_1 = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{|\rr_2-\rr_1|} .\tag{16.2.2} \end{equation}

Because of the superposition principle, the resulting potential is just the sum of the separate potentials due to each charge, and the energy needed to bring in the third charge is just the sum of two terms of the same form as above. Thus, using summation notation, the total energy needed to bring in all of the charges is given by

\begin{equation} U = \sum_{i\lt j} \frac{1}{4\pi\epsilon_0} \frac{q_i q_j}{|\rr_j-\rr_i|} .\tag{16.2.3} \end{equation}

We can express this result in a more useful form by noticing that the expression inside the summation is the same if \(i\) and \(j\) are interchanged. Thus, if we extend the sum to include \(i>j\text{,}\) we pick up the same terms again, resulting in an overall factor of 2. Factoring out the constant, we finally obtain

\begin{equation} U = \frac12 \frac{1}{4\pi\epsilon_0} \sum_{i\ne j} \frac{q_i q_j}{|\rr_j-\rr_i|} .\tag{16.2.4} \end{equation}