Three pont charges q, q & - 2q placed, at the corners of equilateral triangle of side 'L ' Calculate work done by extemal force in moving all the charges far apart without acceleration

Options

A. `(1)/(4 pi epsilon_(0)) (3 q^(2))/(L)`
B. `-(1)/(4 pi epsilon_(0)) (3q^(2))/(L)`
C. `(1)/(4 pi epsilon_(0)) (5q^(2))/(L)`
D. `- (1)/(4 pi epsilon_(0)) (5 q^(2))/(L)`

Answer

To solve the problem of calculating the work done by an external force in moving three point charges \( q, q, \) and \( -2q \) placed at the corners of an equilateral triangle of side \( L \) far apart, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Configuration**:
- We have three charges: \( q_1 = q \), \( q_2 = q \), and \( q_3 = -2q \).
- These charges are placed at the corners of an equilateral triangle with each side of length \( L \).

2. **Calculate Initial Potential Energy**:
- The potential energy \( U \) of a system of point charges is given by the sum of the potential energies of each pair of charges.
- The potential energy between two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by:
\[ U_{12} = k \frac{q_1 q_2}{r} \]
- Here, \( k \) is Coulomb's constant, \( k = \frac{1}{4 \pi \epsilon_0} \).

3. **Calculate Pairwise Potential Energies**:
- For the pairs \( (q_1, q_2) \), \( (q_1, q_3) \), and \( (q_2, q_3) \):
- \( U_{12} = k \frac{q \cdot q}{L} = k \frac{q^2}{L} \)
- \( U_{13} = k \frac{q \cdot (-2q)}{L} = -2k \frac{q^2}{L} \)
- \( U_{23} = k \frac{q \cdot (-2q)}{L} = -2k \frac{q^2}{L} \)

4. **Sum the Potential Energies**:
- The total initial potential energy \( U_{initial} \) of the system is:
\[ U_{initial} = U_{12} + U_{13} + U_{23} \]
\[ U_{initial} = k \frac{q^2}{L} - 2k \frac{q^2}{L} - 2k \frac{q^2}{L} \]
\[ U_{initial} = k \frac{q^2}{L} - 4k \frac{q^2}{L} = -3k \frac{q^2}{L} \]

5. **Final Potential Energy at Infinite Separation**:
- When the charges are moved infinitely far apart, the potential energy \( U_{final} \) becomes zero:
\[ U_{final} = 0 \]

6. **Calculate Work Done by External Force**:
- The work done \( W \) by the external force is equal to the change in potential energy:
\[ W = U_{final} - U_{initial} \]
\[ W = 0 - (-3k \frac{q^2}{L}) = 3k \frac{q^2}{L} \]

7. **Substitute for Coulomb's Constant**:
- If we substitute \( k = \frac{1}{4 \pi \epsilon_0} \):
\[ W = 3 \left(\frac{1}{4 \pi \epsilon_0}\right) \frac{q^2}{L} \]

### Final Answer:
The work done by the external force in moving all the charges far apart without acceleration is:
\[ W = \frac{3q^2}{4 \pi \epsilon_0 L} \]