Two fixed point charges `+4e` and `+e` units are separated by a distance a. Where should a third point charge be placed for it to be in equilibrium?

Explanation

Suppose the three charges are placed as shown in fig. Let the charge q be positive.

For the equilibrium of charge `+q,` we must have Force of repulsion F1 between + 4e and +q = Force of repulsion F2 between + e and +q
or `(1)/(4piepsilon_0)(4exxq)/x^(2)=(1)/(4piepsilon_0)(exxq)/((a-x)^(2))`
or `4(a-x)^(2)=x^(2)`
or `2(a-x)=pmx`
`:.x=(2a)/(3) " or " 2a`
As the charge q is placed between +4e and +e, so only `x = 2a//3` is possible. Hence for equilibrium, the charge q must be placed at a distance 2a/3 from the charge +4e.
We have considered the charge q to be positive.
If we displace it slightly towards charge e, from the equilibrium position, then `F_1` will decrease and `F_2` will increase and a net force `(F_2 – F_1)` will act on q towards left i.e., towards the equilibrium position. Hence the equilibrium of position q is stable. Now if we take charge q to be negative, the force F1 and F2 will be attractive, as shown in fig.

The charge –q will still be in equilibrium at `x = 2a/3.` However, if we displace charge – q slightly towards right, then `F_1` will decrease and `F_2` will increase. A net force `(F_2 – F_1)` will act on –q towards right i.e., away from the equilibrium position. So the equilibrium of the negative q will be unstable.