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).

Explanation

To solve the problem of finding the work done by an external force in increasing the separation of the point charges from \( a \) to \( 2a \), we will follow these steps:

### Step 1: Understand the Initial Configuration
We have three point charges placed at the vertices of an equilateral triangle:
- Charge at vertex A: \( -2q \)
- Charge at vertex B: \( -2q \)
- Charge at vertex C: \( +q \)

The side of the triangle is \( a \).

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

We will calculate the potential energy for each pair of charges:
1. \( U_{AB} \) between charges at A and B:
\[ U_{AB} = k \frac{(-2q)(-2q)}{a} = k \frac{4q^2}{a} \]
2. \( U_{BC} \) between charges at B and C:
\[ U_{BC} = k \frac{(-2q)(+q)}{a} = -k \frac{2q^2}{a} \]
3. \( U_{CA} \) between charges at C and A:
\[ U_{CA} = k \frac{(+q)(-2q)}{a} = -k \frac{2q^2}{a} \]

Now, sum these energies to find the total initial potential energy \( U_i \):
\[ U_i = U_{AB} + U_{BC} + U_{CA} = k \frac{4q^2}{a} - k \frac{2q^2}{a} - k \frac{2q^2}{a} = k \frac{4q^2}{a} - k \frac{4q^2}{a} = 0 \]

### Step 3: Calculate the Final Potential Energy (\( U_f \))
When the charges are moved to a separation of \( 2a \), we repeat the calculation for the new distances:
1. \( U_{AB} \) between charges at A and B:
\[ U_{AB} = k \frac{(-2q)(-2q)}{2a} = k \frac{4q^2}{2a} = k \frac{2q^2}{a} \]
2. \( U_{BC} \) between charges at B and C:
\[ U_{BC} = k \frac{(-2q)(+q)}{2a} = -k \frac{2q^2}{2a} = -k \frac{q^2}{a} \]
3. \( U_{CA} \) between charges at C and A:
\[ U_{CA} = k \frac{(+q)(-2q)}{2a} = -k \frac{2q^2}{2a} = -k \frac{q^2}{a} \]

Now, sum these energies to find the total final potential energy \( U_f \):
\[ U_f = U_{AB} + U_{BC} + U_{CA} = k \frac{2q^2}{a} - k \frac{q^2}{a} - k \frac{q^2}{a} = k \frac{2q^2}{a} - k \frac{2q^2}{a} = 0 \]

### Step 4: Calculate the Work Done
The work done \( W \) by the external force in moving the charges is given by 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.

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