Two equal charges, +Q each, are at a distance r from each other. A third charge is placed on the line joining the above two charges such that all these charges are in equilibrium. Find q, in terms of Q.

Explanation

Suppose the three charges be placed in the manner, as shown in fig.
The charge q will be in equilibrium if the forces exerted on it by the
charges at A and C are equal and opposite.

`k.(Qq)/(x^(2))=k.(Qq)/((r-x)^(2)) " or " x^(2)=(r-x)^(2)`
or `x=r-x " or " x=r/2`
Since the charge at A is repelled by the similar charge at C, so it will be in equilibrium if it is attracted by the charge q at B, i.e., the sign of charge q should be opposite to that of charge Q.
`:.` Force of repulsion between charges at A and C
= Force of attraction between charges at A and B
`k(Q.q)/(r//2)^(2)=k(Q.Q)/(r^2)" or "q=Q/4`