Question

why is the answer b ??how ??
please help me​

Answer 1

The potential energy of a system of two charges is given by

W = k*q₁*q₂/r ........ (1)

Explanation:

Now let's bring the three charges one by one.

Initially the system doesn't contain any charge.

1) Bringing the charge q from infinity to x = (-a). Since there is no other charge present, no force acts over the charge been moved. Thus work done is zero.

2) Moving another charge from infinity to x = 0. A repulsive force acts between the two charges. Work is done and stored in the form of electric potential energy of the system given by (1). So,

work done = k*q*q/a

= k*q²/ a

3) Moving the third charge from infinity to x = a. A repulsive force acts between the three charges. Work is done against two repulsive forces. So,

work done = k*q*q/a + k*q*q/(2a)

= 3*k*q²/(2a)

Now, total work done = k*q²/ a + 3*k*q²/(2a)

= 5*k*q²/(2a)

Substituting for k as 1/4πε•,

Potential energy of the system

= 5*q²/(8πε•a)