Three charges each`20muC` are placed at the corners of an equilateral triangle of side `0.4m` The potential energy of the system is

Options

A. `18xx10^(-6)J`
B. `9J`
C. `9xx10^(-6)J`
D. `27J`

Explanation

To find the potential energy of a system of three charges placed at the corners of an equilateral triangle, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Charges and Distance:**
- Given three charges \( q_1 = q_2 = q_3 = 20 \, \mu C = 20 \times 10^{-6} \, C \).
- The side length of the equilateral triangle \( L = 0.4 \, m \).

2. **Formula for Potential Energy:**
- The potential energy \( U \) of a system of point charges is given by the formula:
\[ U = k \left( \frac{q_1 q_2}{r_{12}} + \frac{q_2 q_3}{r_{23}} + \frac{q_3 q_1}{r_{31}} \right) \]
- Here, \( k \) is Coulomb's constant, \( k = 9 \times 10^9 \, N m^2/C^2 \), and \( r_{ij} \) is the distance between charges \( q_i \) and \( q_j \).

3. **Calculate Each Pair's Contribution:**
- Since all charges are equal and the distances are the same:
\[ U = k \left( \frac{q^2}{L} + \frac{q^2}{L} + \frac{q^2}{L} \right) = 3 \cdot \frac{k q^2}{L} \]
- Substitute \( q = 20 \times 10^{-6} \, C \) and \( L = 0.4 \, m \):
\[ U = 3 \cdot \frac{9 \times 10^9 \cdot (20 \times 10^{-6})^2}{0.4} \]

4. **Calculate \( q^2 \):**
- Calculate \( (20 \times 10^{-6})^2 = 400 \times 10^{-12} \, C^2 \).

5. **Substitute Values:**
- Now substitute back into the equation:
\[ U = 3 \cdot \frac{9 \times 10^9 \cdot 400 \times 10^{-12}}{0.4} \]

6. **Simplify the Expression:**
- Calculate the numerator:
\[ 9 \times 10^9 \cdot 400 \times 10^{-12} = 3600 \times 10^{-3} = 3.6 \, J \]
- Now divide by \( 0.4 \):
\[ U = 3 \cdot \frac{3.6}{0.4} = 3 \cdot 9 = 27 \, J \]

7. **Final Result:**
- The potential energy of the system is:
\[ U = 27 \, J \]