A charge $Q$ is placed at each of the opposite corners of a square. A charge $q$ is placed at each of the other two corners. If the net electrical force on $Q$ is zero, then $\frac{Q}{q}$ equals:

• A

  $-1$
• B

  $1$
• C

  $-\frac{1}{\sqrt{2}}$
• Correct answer
  D

  $-2 \sqrt{2}$

Solution

It is given that $\vec{F}_{A}+\vec{F}_{B}+\vec{F}_{D}=0$ Where $\vec{F}_{A}, \vec{F}_{B}$ and $\vec{F}_{D}$ are the forces applied by charges placed at $A, B$ and $D$ on the charge placed at $C$. $\begin{array}{l} \Rightarrow \overrightarrow{\mathrm{F}}_{\mathrm{B}}+\overrightarrow{\mathrm{F}}_{\mathrm{D}}=-\overrightarrow{\mathrm{F}}_{\mathrm{A}} \\ \left|\overrightarrow{\mathrm{F}}_{\mathrm{B}}+\overrightarrow{\mathrm{F}}_{\mathrm{D}}\right|=\sqrt{2} \frac{\mathrm{KqQ}}{\mathrm{a}^{2}} \\ \Rightarrow\left|\overrightarrow{\mathrm{F}}_{\mathrm{A}}\right|=\frac{\mathrm{KQ}^{2}}{2 a^{2}} \\ \therefore \sqrt{2} \frac{\mathrm{KqQ}}{a^{2}}=-\frac{\mathrm{KQ}^{2}}{2 a^{2}} \\ \Rightarrow \frac{\mathrm{Q}}{q}=-2 \sqrt{2} \end{array}$

Hence, the correct answer is option: 'D'.