Three point charges 1C, -2C and -2C are placed at the vertices of an equilateral triangle of side length one meter. The work done by an external force to increase the separation of the charges to 2 meters in joules is: (`in_(0)` is permittivity of air)

Options

A. `(1)/(4pi in_(0))`
B. `(1)/(8pi in_(0))`
C. `(1)/(16pi in_(0))`
D. 0

Explanation

To solve the problem of calculating the work done by an external force to increase the separation of the charges from 1 meter to 2 meters, we will follow these steps:

### Step 1: Identify the Charges and Their Configuration
We have three point charges:
- Charge \( q_1 = 1 \, C \)
- Charge \( q_2 = -2 \, C \)
- Charge \( q_3 = -2 \, C \)

These charges are placed at the vertices of an equilateral triangle with a side length of 1 meter.

### Step 2: Calculate Initial Potential Energy
The potential energy \( U \) between two point charges is given by the formula:
\[ U = k \frac{q_1 q_2}{r} \]
where \( k = \frac{1}{4 \pi \epsilon_0} \) and \( r \) is the distance between the charges.

#### Calculate Pairwise Potential Energies:
1. Between \( q_1 \) and \( q_2 \):
\[ U_{12} = k \frac{(1)(-2)}{1} = -2k \]

2. Between \( q_1 \) and \( q_3 \):
\[ U_{13} = k \frac{(1)(-2)}{1} = -2k \]

3. Between \( q_2 \) and \( q_3 \):
\[ U_{23} = k \frac{(-2)(-2)}{1} = 4k \]

#### Total Initial Potential Energy:
\[ U_{initial} = U_{12} + U_{13} + U_{23} = -2k - 2k + 4k = 0 \]

### Step 3: Calculate Final Potential Energy
Now, we need to calculate the potential energy when the charges are separated to a distance of 2 meters.

1. Between \( q_1 \) and \( q_2 \):
\[ U'_{12} = k \frac{(1)(-2)}{2} = -k \]

2. Between \( q_1 \) and \( q_3 \):
\[ U'_{13} = k \frac{(1)(-2)}{2} = -k \]

3. Between \( q_2 \) and \( q_3 \):
\[ U'_{23} = k \frac{(-2)(-2)}{2} = 2k \]

#### Total Final Potential Energy:
\[ U_{final} = U'_{12} + U'_{13} + U'_{23} = -k - k + 2k = 0 \]

### Step 4: Calculate Work Done
The work done \( W \) by the external force is equal to the negative change in potential energy:
\[ W = - (U_{final} - U_{initial}) = - (0 - 0) = 0 \]

### Final Answer
The work done by an external force to increase the separation of the charges to 2 meters is:
\[ \boxed{0 \, \text{Joules}} \]