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 total electrostatic potential energy \(U\) of a system of two point charges \(q_1\) and \(q_2\) located at positions \(\vec{r}_1\) and \(\vec{r}_2\) in an external electric field \(\vec{E}\) consists of three parts: the work done to bring \(q_1\) into the external field, the work done to bring \(q_2\) into the external field, and the mutual electrostatic interaction energy between \(q_1\) and \(q_2\): \[ U = q_1 V(\vec{r}_1) + q_2 V(\vec{r}_2) + \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r_{12}} \] where \(V(\vec{r})\) is the external electric potential, obtained by integrating the electric field: \(V(x) = -\int \vec{E} \cdot d\vec{x}\).

Step 1: Convert all given parameters to SI units.
Ensure consistency in numerical calculations by transforming centimeters to meters and microcoulombs to coulombs.
- Charge \(q_1 = -2\,\mu\text{C} = -2 \times 10^{-6}\,\text{C}\) at position \(x_1 = -30\,\text{cm} = -0.3\,\text{m}\).
- Charge \(q_2 = 5\,\mu\text{C} = 5 \times 10^{-6}\,\text{C}\) at position \(x_2 = 30\,\text{cm} = 0.3\,\text{m}\).
- Constant \(A = 9 \times 10^5\,\text{N}\cdot\text{m}^2/\text{C}\).
- The separation distance between the two charges is: \[ r_{12} = |x_2 - x_1| = |0.3 - (-0.3)| = 0.6\,\text{m} \]

Step 2: Find the expression for the external electric potential \(V(x)\).
Integrate the electric field function along the x-axis.
Given \(\vec{E} = \frac{A}{x^2}\hat{i}\), the potential function \(V(x)\) relative to infinity is: \[ V(x) = -\int \frac{A}{x^2} dx = -A \left( -\frac{1}{x} \right) = \frac{A}{x} \] Now evaluate the potential at the specific coordinate positions of both charges: \[ V(x_1) = V(-0.3) = \frac{9 \times 10^5}{-0.3} = -3 \times 10^6\,\text{V} \] \[ V(x_2) = V(0.3) = \frac{9 \times 10^5}{0.3} = 3 \times 10^6\,\text{V} \]

Step 3: Calculate each term of the potential energy formula.
Compute individual energetic components precisely.
1. Interaction energy with the external field for \(q_1\): \[ U_1 = q_1 V(x_1) = (-2 \times 10^{-6}\,\text{C}) \times (-3 \times 10^6\,\text{V}) = 6\,\text{J} \] 2. Interaction energy with the external field for \(q_2\): \[ U_2 = q_2 V(x_2) = (5 \times 10^{-6}\,\text{C}) \times (3 \times 10^6\,\text{V}) = 15\,\text{J} \] 3. Mutual potential energy between \(q_1\) and \(q_2\): \[ U_{12} = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r_{12}} = (9 \times 10^9) \times \frac{(-2 \times 10^{-6}) \times (5 \times 10^{-6})}{0.6} \] \[ U_{12} = (9 \times 10^9) \times \frac{-10 \times 10^{-12}}{0.6} = \frac{-90}{0.6} = -150\,\text{J} \]

Step 4: Sum all components to find total potential energy.
\[ U = U_1 + U_2 + U_{12} = 6 + 15 - 150 = 21 - 150 = -129\,\text{J} \] *(Note: If calculating using absolute values of coordinates for potential magnitude relative to the source origin, let's re-verify matching the target option formats if \(V = \frac{A}{|x|}\) is taken: \(U = -6 + 15 - 150 = -141\,\text{J}\). Let's review standard context values. If \(A\) is extraordinarily large or scaled differently in specific book definitions, we choose based on exact algebraic substitution.)*