Question

Two equal point charges $Q=\sqrt{2}\mu C$ are placed at each of the two opposite corners of a square and equal point -charges $q$ at each of the other two corner.What must be the value of $q$ so that the resultant force on $Q$ is zero?

Answer 1

Let $a$ metre the side of the side of the square.The length of a diagonal will be $\sqrt{2}a$.The force on the positive charge $Q$ at A due to the positive charge $Q$ at $C$ is $F=\frac{1}{4\pi {\epsilon }_0}\frac{Q^2}{2a^2}$ and is directed outward,as shown.In order to make net force on $Q$ zero,the resultant of the other two force $F_1$ and $F_2$ on Q due to the charges $q$ and $q$ should be equal and opposite to $F$.Clearly $F_1$ and $F_2$ will be directed as shown and for this both $q^{\prime }$ should be negative.
now
$F_1=F_2=\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{a^2}$
since $F_1$ and $F_2$ are at right angles, their resultant is
$F_{12}=\frac{1}{4\pi {\epsilon }_0}\frac{\sqrt{2}Qq}{a^2}$
For the equilibrium of $Q$, we musst have
$F=F_{12}$
But $Q+\sqrt{2}\mu C$
$\therefore q=-\frac{\sqrt{2}\mu C}{2\sqrt{2}}=-0.5\mu C$