Question

Three point charges $+q$, $+q$, and $-q$ are placed at the vertices of an equilateral triangle of side length $L$. The net electrostatic potential energy of this system of charges is given by:

• A. \( \frac{1}{4\pi\varepsilon_0} \frac{q^2}{L} \)
• B. \( -\frac{1}{4\pi\varepsilon_0} \frac{q^2}{L} \)
• C. \( -\frac{3}{4\pi\varepsilon_0} \frac{q^2}{L} \)
• D. \( 0 \)

Hint:

Potential energy is a scalar quantity, not a vector! Never split it into horizontal or vertical components. Simply carry the sign of the charges through the equations and sum them up using basic arithmetic.

Answer 1

The Correct Option is B

Concept: The electrostatic potential energy of a collection of stationary point charges represents the total external work done to bring these charges from an infinite separation to their specific coordinates in space without accelerating them. Based on the fact that electrostatic forces are conservative, the potential energy depends strictly on the final configuration, not the path taken. For a pair of point charges \(q_i\) and \(q_j\) separated by a spatial distance \(r_{ij}\), the mutual potential energy is a scalar value given by Coulomb's law derivation: \[ U_{ij} = \frac{1}{4\pi\varepsilon_0} \frac{q_i q_j}{r_{ij}} \] When dealing with a system containing more than two charges, the net electrostatic potential energy is computed by taking the algebraic sum of the potential energies of every single unique pair configuration. For a system of three charges, this translates to: \[ U_{\text{total}} = U_{12} + U_{23} + U_{13} \]

Step 1: Identify the individual charges and their geometric separations. Let us map out our three point charges sitting at the corners of the equilateral triangle:
• \(q_1 = +q\)
• \(q_2 = +q\)
• \(q_3 = -q\) Since they occupy the vertices of an equilateral triangle, every single pair is separated by the exact same side length, meaning: \[ r_{12} = r_{23} = r_{13} = L \]

Step 2: Calculate the scalar interaction energy for each individual pair. Let us write out the explicit energy expression for each pair, keeping careful track of the algebraic signs: 1. Interaction between charge 1 and charge 2 (both positive): \[ U_{12} = \frac{1}{4\pi\varepsilon_0} \frac{(+q)(+q)}{L} = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{L} \] 2. Interaction between charge 2 and charge 3 (one positive, one negative): \[ U_{23} = \frac{1}{4\pi\varepsilon_0} \frac{(+q)(-q)}{L} = -\frac{1}{4\pi\varepsilon_0} \frac{q^2}{L} \] 3. Interaction between charge 1 and charge 3 (one positive, one negative): \[ U_{13} = \frac{1}{4\pi\varepsilon_0} \frac{(+q)(-q)}{L} = -\frac{1}{4\pi\varepsilon_0} \frac{q^2}{L} \]

Step 3: Algebraically sum up the pair energies to find the net system energy. Now, add the three scalar parts together: \[ U_{\text{total}} = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q^2}{L} + \left(-\frac{q^2}{L}\right) + \left(-\frac{q^2}{L}\right) \right] \] Combining the common terms inside the bracket: \[ U_{\text{total}} = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q^2}{L} - \frac{q^2}{L} - \frac{q^2}{L} \right] \] \[ U_{\text{total}} = -\frac{1}{4\pi\varepsilon_0} \frac{q^2}{L} \] The negative sign indicates that the system is bound; external work must be put into the configuration to completely pull these charges infinitely far apart from each other.