Question

Two point charges \(-2\,\mu\text{C}\) and \(5\,\mu\text{C}\) are placed at \((-30\,\text{cm}, 0)\) and \((30\,\text{cm}, 0)\) respectively in an external electric field \(\vec{E} = \frac{A}{x^2} \hat{i}\), where \(A = 9 \times 10^5\,\text{N}\cdot\text{m}^2\cdot\text{C}^{-1}\). Find the electrostatic potential energy of this configuration.

Hint:

Always remember that when an external field is present, never forget to calculate the work required to set up individual charges inside that field alongside their mutual system interaction energy: \(U_{\text{total}} = U_{\text{ext}} + U_{\text{mutual}}\).

Answer 1

Concept: The potential energy of two point charges placed in an external field is the sum of the energy of each charge in the external field and the mutual interaction energy of the two charges: \[ U = q_1 V(x_1) + q_2 V(x_2) + \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r_{12}} \] where the external potential is obtained from \(V(x) = -\int E\,dx\).

Step 1: Data in SI units.
\(q_1 = -2\,\mu\text{C} = -2\times10^{-6}\,\text{C}\) at \(x_1 = -0.3\,\text{m}\); \(q_2 = 5\,\mu\text{C} = 5\times10^{-6}\,\text{C}\) at \(x_2 = +0.3\,\text{m}\); \(A = 9\times10^{5}\,\text{N m}^2\,\text{C}^{-1}\); separation \(r_{12} = 0.6\,\text{m}\).

Step 2: External potential.
For \(E = \frac{A}{x^2}\): \[ V(x) = -\int \frac{A}{x^2}\,dx = \frac{A}{x} \] \[ V(x_1) = \frac{9\times10^{5}}{-0.3} = -3\times10^{6}\,\text{V}, \qquad V(x_2) = \frac{9\times10^{5}}{0.3} = 3\times10^{6}\,\text{V} \]

Step 3: Energy of each charge in the external field.
\[ U_1 = q_1 V(x_1) = (-2\times10^{-6})(-3\times10^{6}) = 6\,\text{J} \] \[ U_2 = q_2 V(x_2) = (5\times10^{-6})(3\times10^{6}) = 15\,\text{J} \]

Step 4: Mutual interaction energy.
\[ U_{12} = \frac{(9\times10^{9})(-2\times10^{-6})(5\times10^{-6})}{0.6} = \frac{-9\times10^{-2}}{0.6} = -0.15\,\text{J} \]

Step 5: Total.
\[ U = 6 + 15 - 0.15 = 20.85\,\text{J} \]
Final Answer: The electrostatic potential energy of the configuration is \(U = 20.85\,\text{J}\) (about 21 J).