Question

Three equal electric charges of each charge \( q \) are placed at the vertices of an equilateral triangle of side length \( L \), then potential energy of the system is:

• A. \( \frac{1}{4\pi\epsilon_0} \frac{3q^2}{L} \)
• B. \( \frac{1}{4\pi\epsilon_0} \frac{q^2}{3L} \)
• C. \( \frac{1}{4\pi\epsilon_0} \frac{2q^2}{3L} \)
• D. \( \frac{1}{4\pi\epsilon_0} \frac{q^2}{L} \)

Hint:

For a system of point charges, the total potential energy is the sum of the potential energies of all pairs of charges. When the charges are placed symmetrically, such as in an equilateral triangle, this simplifies the calculation.

Answer 1

The Correct Option is A

We are given three equal electric charges of charge \( q \), placed at the vertices of an equilateral triangle with side length \( L \). The task is to find the potential energy of this system.

Step 1: The potential energy \( U \) of a system of charges is given by the formula:

where \( k = \frac{1}{4\pi\epsilon_0} \), \( q_i \) and \( q_j \) are the magnitudes of the charges, and \( r_{ij} \) is the distance between the charges.

Step 2: For this system, all three charges are of equal magnitude \( q \) and the distance between any two charges is \( L \). Therefore, the potential energy of the system is the sum of the potential energies due to each pair of charges:

\( U = \frac{k q^2}{L} + \frac{k q^2}{L} + \frac{k q^2}{L} = 3 \times \frac{k q^2}{L} \)

Substituting \( k = \frac{1}{4\pi\epsilon_0} \), we get:

\( U = \frac{1}{4\pi\epsilon_0} \frac{3q^2}{L} \)

Thus, the potential energy of the system is \( \frac{1}{4\pi\epsilon_0} \frac{3q^2}{L} \).