Three point charges q, -2q and -2q are placed at the vertices of an equilateral triangle of side a. Find the work done by external force to increase their separation to 2a (in joules).

To find the work done by an external force to increase the separation of the three point charges from a distance of \( a \) to \( 2a \), we first need to calculate the initial and final potential energy of the system.

Initial Potential Energy

The potential energy \( U \) between two point charges is given by the formula:

U = k * (q1 * q2) / r

where \( k \) is Coulomb's constant, \( q1 \) and \( q2 \) are the charges, and \( r \) is the distance between them.

Calculating Initial Energies

• For charges \( q \) and \( -2q \):

Distance = \( a \)

U1 = k * (q * -2q) / a = -2kq² / a

• For charges \( -2q \) and \( -2q \):

Distance = \( a \)

U2 = k * (-2q * -2q) / a = 4kq² / a

• For charges \( q \) and \( -2q \) (the other pair):

Distance = \( a \)

U3 = -2kq² / a

The total initial potential energy \( U_{initial} \) is:

U_{initial} = U1 + U2 + U3 = -2kq²/a + 4kq²/a - 2kq²/a = 0

Final Potential Energy

Now, we calculate the potential energy when the charges are separated to a distance of \( 2a \).

Calculating Final Energies

• For charges \( q \) and \( -2q \):

Distance = \( 2a \)

U1' = k * (q * -2q) / (2a) = -kq² / a

• For charges \( -2q \) and \( -2q \):

Distance = \( 2a \)

U2' = k * (-2q * -2q) / (2a) = 2kq² / a

• For charges \( q \) and \( -2q \) (the other pair):

Distance = \( 2a \)

U3' = -kq² / a

The total final potential energy \( U_{final} \) is:

U_{final} = U1' + U2' + U3' = -kq²/a + 2kq²/a - kq²/a = 0

Work Done by External Force

The work done \( W \) by the external force is the change in potential energy:

W = U_{final} - U_{initial} = 0 - 0 = 0

Thus, the work done by the external force to increase the separation of the charges to \( 2a \) is:

W = 0 joules