Four charge `Q, q, Q` and q are kept at the four corners of a square as shown below. What is the relation between Q and q, so that the net force on a charge q is zero ?

Explanation

Here both the .q. will have same sign either positive or negative. Similarly both the Q will have same sign. Let us make the force on upper right corner q equal to zero.
Lower .q. will apply a repelling force `F_(1)` on upper q because both the charges have same sign. To balance this force both .Q. must apply attractive force `vec(F_(2))` and `vec(F_(3))` of equal magnitude, hence, Q and q will have opposite signs. Now the resultant of `vec(F_(2))` and `vec(F_(3))` will be `F sqrt(2)` (Parellelogram law of vector addition) if `|vec(F_(2))|=|vec(F_(3))|=F`. Also note that `F sqrt(2)` will be exactly equal and opposite to `F_(1)`.
So, `F_(1)=F sqrt(2)`
From, Coulomb.s law,
`F_(1)=(kq^(2))/((d sqrt(2))^(2))" and "F=(kQq)/(d^(2))`
`because F_(2)=sqrt(2) F`
`:." "(q^(2))/((d sqrt(2))^(2))=(sqrt(2)Qq)/(d^(2))`
`:." "Q=(q)/(2sqrt(2))`
But as we said Q and q have opposite sign so, `q=-2sqrt(2)Q`.