Question

A charge $+q$ is fixed to each of three corners of a square. On the empty corner a charge $Q$ is placed such that there is no net electrostatic force acting on the diagonally opposite charge. Then:

• A. $Q=-2q$
• B. $Q=-2\sqrt{2}q$
• C. $Q=-\sqrt{2}q$
• D. $Q=-4q$

Answer 1

The correct option is B $Q=-2\sqrt{2}q$

Charges $+q$ are placed at A, C and D. And charge $Q$ is place at B.

Now, it is given that net force on charge $q$ diagonally to $Q$ that is at $D$ is 0.

From figure, $F_1=F_3=\frac{q^2}{4\pi {ϵ}_o{a}^2}$

The resultant of $F_1$ and $F_3$ acting perpendicular is along diagonal BD and is equal to $\sqrt{2}{F}_1=\frac{\sqrt{2}{q}^2}{4\pi {ϵ}_o{a}^2}$

$BD=\sqrt{2}a$

And, force on $+q$ at D due to $Q$, $F_2=\frac{qQ}{4\pi {ϵ}_{o}B{D}^2}$

$⟹F_2=\frac{qQ}{8\pi {ϵ}_o{a}^2}$ acting along Bd.

Hence, net force $F=\sqrt{2}{F}_1+F_2=0$ along BD

$⟹\frac{\sqrt{2}{q}^2}{4\pi {ϵ}_o{a}^2}+\frac{qQ}{8\pi {ϵ}_o{a}^2}=0$

$⟹Q=-2\sqrt{2}q$

Answer-(B)