Example

Question:

(a) Determine the electrostatic potential energy of a system consisting of two charges 7 μC and -2 μC (and with no external field) placed at (-9 cm, 0, 0) and (9 cm, 0, 0) respectively.
(b) How much work is required to separate the two charges infinitely away from each other?
(c) Suppose that the same system of charges is now placed in an external electric field \(E = A(1/r^2)\); \(A = 9 \times 10^5 \, \text{NC}^{-1} \, \text{m}^2\). What would the electrostatic energy of the configuration be?

Solution:

(a) \[ U = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r} = 9 \times 10^9 \times \frac{7 \times (-2) \times 10^{-12}}{0.18} = -0.7\,\text{J} \] (b) \[ W = U_2 - U_1 = 0 - (-0.7) = 0.7\,\text{J} \] (c)
The mutual interaction energy remains unchanged. The energy of interaction of the two charges with the external field gives \[ q_1 V(\mathbf{r}_1) + q_2 V(\mathbf{r}_2) = \frac{A \cdot 7 \mu \text{C}}{0.09\,\text{m}} + \frac{A \cdot (-2) \mu \text{C}}{0.09\,\text{m}} \] The net electrostatic energy is \[ q_1 V(\mathbf{r}_1) + q_2 V(\mathbf{r}_2) + \frac{q_1 q_2}{4 \pi \varepsilon_0 r_{12}} \] \[ = \frac{A \cdot 7\mu\text{C}}{0.09\,\text{m}} + \frac{A \cdot (-2)\mu\text{C}}{0.09\,\text{m}} - 0.7\,\text{J} = 70 - 20 - 0.7 = 49.3\,\text{J} \]