Question

A charge Q is placed on to two opposite corner of a square. A charge q is placed at each of other two corners. Given that resultant electric force on Q is zero, then Q is equal to :

• A. $\left(2\sqrt{2}\right)/q$
• B. $-q/\left(2\sqrt{2}\right)$
• C. $\left(2\sqrt{2}\right)q$
• D. $(-2\sqrt{2})q$

Answer 1

The correct option is D $(-2\sqrt{2})q$

The force in between two charges Q and q is $\overrightarrow{F}=k\frac{Qq}{r^2}\hat{r}=k\frac{Qq}{r^3}\overrightarrow{r}$ where $r$ is the distance between them.

Here the resultant force on Q on the side of the square is

$F_R=F_q+F_Q+F_q=k\frac{Qq}{a^3}\left(a\hat{i}\right)+k\frac{Q^2}{(\sqrt{2}a{)}^3}(a\hat{i}+a\hat{j})+k\frac{Qq}{a^3}\left(a\hat{j}\right)=k\frac{Qq}{a^3}(a\hat{i}+a\hat{j})+k\frac{Q^2}{(\sqrt{2}a{)}^3}(a\hat{i}+a\hat{j})$

since $F_R=0\Rightarrow \frac{Qq}{a^3}=-\frac{Q^2}{(\sqrt{2}a{)}^3}$

or $Q=(-2\sqrt{2})q$