Question

Two point charges $-2\mu C$ and $5\mu C$ 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 $A = 9 \times 10^5 \text{ Nm}^2\text{C}^{-1}$. Find the electrostatic potential energy of this configuration.

Hint:

Never casually ignore the critical algebraic signs of charges when calculating potential energy, as they directly dictate whether the energy actively contributes positively or formally acts as a binding negative factor.
Ensure all spatial distances are appropriately converted strictly to standard meters before invoking SI constant calculations.

Answer 1

Concept:
• The total electrostatic potential energy of a multi-charge system situated within a pre-existing external electric field is a scalar sum of multiple independent energy components.

• First, it fundamentally includes the individual physical work done in securely bringing each charge from infinity to its spatial position within the external field ($U = qV$).

• Secondly, it includes the mutual interaction potential energy exclusively existing between the localized charges themselves, computed using Coulomb's electrostatic potential formula.

Step 1: Determine the External Electric Potential Function
The external electric field is mathematically given strictly as $\vec{E} = \frac{A}{x^2} \hat{i}$.
The scalar electric potential $V(x)$ intricately and universally relates to the electric field via the line integral equation:
\[ V(x) = -\int E dx \]
Integrating the provided field function perfectly from infinity yields the exact potential function:
\[ V(x) = -\int \frac{A}{x^2} dx = -\left( -\frac{A}{x} \right) = \frac{A}{x} \]

Step 2: Calculate Individual Energies of Charges in the External Field
We logically compute the exact potential strictly at the spatial position of the first charge, $x_1 = -30 \text{ cm} = -0.3 \text{ m}$:
\[ V_1 = \frac{9 \times 10^5}{-0.3} = -3 \times 10^6 \text{ V} \]
The energy of the first charge $q_1 = -2 \mu C = -2 \times 10^{-6} \text{ C}$ inside this localized field is:
\[ U_1 = q_1 V_1 = (-2 \times 10^{-6}) \times (-3 \times 10^6) = 6 \text{ J} \]
Similarly, systematically compute the potential exactly at the position of the second charge, $x_2 = 30 \text{ cm} = 0.3 \text{ m}$:
\[ V_2 = \frac{9 \times 10^5}{0.3} = 3 \times 10^6 \text{ V} \]
The energy of the second charge $q_2 = 5 \mu C = 5 \times 10^{-6} \text{ C}$ securely inside this localized field is:
\[ U_2 = q_2 V_2 = (5 \times 10^{-6}) \times (3 \times 10^6) = 15 \text{ J} \]

Step 3: Calculate Mutual Interaction Energy Between the Charges
The absolute spatial distance separating the two fixed point charges is $r = |0.3 - (-0.3)| = 0.6 \text{ m}$.
The mutual electrostatic potential energy is computed explicitly by Coulomb's two-charge potential formula:
\[ U_{12} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r} \]
Substitute the universally known standard constant and the exact charge values carefully:
\[ U_{12} = 9 \times 10^9 \frac{(-2 \times 10^{-6}) \times (5 \times 10^{-6})}{0.6} \]
\[ U_{12} = \frac{-90 \times 10^{-3}}{0.6} = -150 \times 10^{-3} \text{ J} = -0.15 \text{ J} \]

Step 4: Calculate Total Electrostatic Potential Energy
The absolute total energy $U_{total}$ of the entire configuration is the direct scalar sum of all three previously calculated independent energy terms:
\[ U_{total} = U_1 + U_2 + U_{12} \]
\[ U_{total} = 6 + 15 - 0.15 = 20.85 \text{ J} \]
The final computed potential energy of the comprehensive system is strictly 20.85 Joules.