The point charges `-2q, -2q` and `+q` are put on the vertices of an equilateral triangle of side `a`. Find the work done by some external force in increasing the separation to `2a` (in joules).

Options

A. `(1)/(4piepsilon_(0)).(2q^(2))/(a)`
B. negativee
C. zero
D. `(1)/(4piepsilon_(0)).(3q^(2))/(a)`

Answer

To find the work done by an external force in increasing the separation of the charges from a distance `a` to `2a`, we will calculate the initial and final potential energy of the system of charges and then determine the work done based on the change in potential energy.

### Step-by-Step Solution:

1. **Identify the Charges and Their Positions**:
We have three charges placed at the vertices of an equilateral triangle:
- Charge 1: \( Q_1 = -2q \)
- Charge 2: \( Q_2 = -2q \)
- Charge 3: \( Q_3 = +q \)

2. **Calculate the Initial Potential Energy**:
The initial separation between the charges is \( a \). The potential energy \( U \) between two point charges is given by:
\[ U = k \frac{Q_1 Q_2}{r} \]
where \( k \) is Coulomb's constant and \( r \) is the distance between the charges.

- Potential energy between \( Q_1 \) and \( Q_2 \):
\[ U_{12} = k \frac{(-2q)(-2q)}{a} = k \frac{4q^2}{a} \]

- Potential energy between \( Q_1 \) and \( Q_3 \):
\[ U_{13} = k \frac{(-2q)(+q)}{a} = -k \frac{2q^2}{a} \]

- Potential energy between \( Q_2 \) and \( Q_3 \):
\[ U_{23} = k \frac{(-2q)(+q)}{a} = -k \frac{2q^2}{a} \]

Now, summing these potential energies gives the total initial potential energy \( U_i \):
\[ U_i = U_{12} + U_{13} + U_{23} = k \frac{4q^2}{a} - k \frac{2q^2}{a} - k \frac{2q^2}{a} = 0 \]

3. **Calculate the Final Potential Energy**:
Now, we increase the separation to \( 2a \). We will recalculate the potential energies with the new distance.

- Potential energy between \( Q_1 \) and \( Q_2 \):
\[ U_{12}' = k \frac{(-2q)(-2q)}{2a} = k \frac{4q^2}{2a} = k \frac{2q^2}{a} \]

- Potential energy between \( Q_1 \) and \( Q_3 \):
\[ U_{13}' = k \frac{(-2q)(+q)}{2a} = -k \frac{2q^2}{2a} = -k \frac{q^2}{a} \]

- Potential energy between \( Q_2 \) and \( Q_3 \):
\[ U_{23}' = k \frac{(-2q)(+q)}{2a} = -k \frac{2q^2}{2a} = -k \frac{q^2}{a} \]

Now, summing these potential energies gives the total final potential energy \( U_f \):
\[ U_f = U_{12}' + U_{13}' + U_{23}' = k \frac{2q^2}{a} - k \frac{q^2}{a} - k \frac{q^2}{a} = 0 \]

4. **Calculate the Work Done**:
The work done \( W \) by the external force is equal to the change in potential energy:
\[ W = U_f - U_i = 0 - 0 = 0 \, \text{Joules} \]

### Conclusion:
The work done by the external force in increasing the separation of the charges from \( a \) to \( 2a \) is \( 0 \) Joules.