Question

Three point charges each carrying a charge \(q\) are placed on the vertices of an equilateral triangle of side \(L\). The electrostatic potential energy of the configuration is:

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

Hint:

For \(N\) point charges, always sum the potential energy over all distinct pairs using \(U = \sum_{i<j} \dfrac{1}{4\pi \varepsilon_0} \dfrac{q_i q_j}{r_{ij}}\). Symmetry often simplifies the calculation.

Answer 1

The Correct Option is C

Step 1: Understanding the concept.
The electrostatic potential energy of a system of charges is the sum of the potential energies for all distinct pairs of charges. For two point charges \(q_1\) and \(q_2\) separated by a distance \(r\), the potential energy is \[ U = \frac{1}{4\pi \varepsilon_0} \frac{q_1 q_2}{r} \]

Step 2: Apply to the given system.
In an equilateral triangle, all three sides are equal to \(L\). There are 3 distinct pairs of charges: (1,2), (2,3), and (3,1). Hence, the total potential energy is the sum of potential energies of these three pairs: \[ U = 3 \times \frac{1}{4\pi \varepsilon_0} \frac{q^2}{L} \]

Step 3: Simplify the expression.
\[ U = \frac{3}{4\pi \varepsilon_0} \frac{q^2}{L} \]

Step 4: Conclusion.
The electrostatic potential energy of three identical charges at the vertices of an equilateral triangle is \(\dfrac{3q^2}{4\pi \varepsilon_0 L}\).