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

Answer 2

To find the potential energy of the system of three equal charges placed at the vertices of an equilateral triangle, we need to consider the potential energy due to each pair of charges. The formula for the potential energy \( U \) between two point charges is given by:

\[ U = \frac{1}{4\pi\epsilon_0} \frac{q_1q_2}{r} \]

where \( q_1 \) and \( q_2 \) are the point charges and \( r \) is the distance between them.

In this case, each charge \( q \) interacts with the other two charges, forming three pairs. Since the triangle is equilateral with side length \( L \), the potential energy contribution from each pair is:

\[ U_{\text{pair}} = \frac{1}{4\pi\epsilon_0} \frac{q^2}{L} \]

Since there are three pairs in the system, we add the potential energy of all pairs:

\[ U_{\text{total}} = 3 \times \frac{1}{4\pi\epsilon_0} \frac{q^2}{L} = \frac{3q^2}{4\pi\epsilon_0 L} \]

This gives us the total potential energy of the system:

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

The correct answer from the provided options is:

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