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

Explanation

The charges are `q_1 = +4q and q_2= +q`
Distance between the charges = d
Force between the charges
`F= 1/(4piepsilon_0)(q_1q_2)/r^2`

AB = d, AC = q, CB = d-x
Let a charge q be placed between two charges. Force on the charge C exerted by A is given by `F_1= (q(4q))/(4piepsilon_0x^2)` directed away from A )4q)/ Force on the charge C exerted by B is given by `F_2= (q(q))/(4piepsilon_0(d-x)^2)` directed away from B (q).
For the charge q to be in equilibrium,
`F_1=F_2`
`(q(4q))/(4piepsilon_0x^2)= (q(q))/(4piepsilon_0(d-x)^2)`
`4/x^2= 1/(d-x)^2`
`2/x= 1/(d-x)`
2d - 2x = x
2d = 3x
`:.x= (2d)/3`
The charge q should be placed at a distance `(2d)/3` from charge 4q.