Two particles (free to move) with charges `+q` and `+4q` are a distance L apart. A third charge is placed so that the entire system is in equilibrium.
(a) Find the location, magnitude and sign of the third charge.
(b) Show that the equilibrium is unstable.

Answer

A, B, C, D

Explanation

`x+y=L`.i
Let force on -Q charge should be zero.
`:. (kQ.q)/x^2=(kQ(4q))/y^2`
`:. x/y=1/2`…….ii
`From eqn i and ii we get
`x=L/3` and `y=(2L)/3`
Net force on 4q should be zero.
`:. (k|Q|(4q))/((2L//3)^2)=(kq(4q))/L^2`
`:. |Q|=4/9q`
With sign,
`Q=-4/9q`
Similarly net force on q should be zero.
`:. (k|Q|q)/((L//3)^2)=(kq(4q))/L^2 or |Q|=4/9q `
If we, sightly displace `-Q` towards `4q`, attraction between these charges will increase, hence `-Q` will move towards `+4q` and it will not return back. Hence equilibrium is unstable.