Explanation :
It is given that, q and 4q are placed at a distance of l.
Let x is the distance where third charge is placed so that the entire three charge system is in static equilibrium.
Equilibrium means the net force acting on the system of charges is equal to zero. Let Q is the third charge.
So, [tex]\dfrac{kqQ}{x^2}=\dfrac{k(Q)(4q)}{(l-x)^2}[/tex]
On solving,
[tex]x=\dfrac{l}{3}[/tex]
For magnitude :
[tex]\dfrac{kqQ}{(l/3)^2}=\dfrac{kq4q}{l^2}[/tex]
using [tex]x=\dfrac{l}{3}[/tex]
[tex]Q=\dfrac{4q}{9}[/tex]
So, a charge of -4q/9 is at a distance of l/3 is placed. It is placed to the right of +q.
