Question

Charges are placed at the corners of a square of side $a$ , as shown in the following figure. The charged particle placed at $A$ is in equilibrium. The ratio $\frac{q_1}{q_2}$ is

• A. $1$
• B. $2\sqrt{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{3}{\sqrt{2}}$

Answer 1

Answer: (B)

The charge at $A$ experiences three forces, $F_1$ , $F_2$ and $F_3$ , as shown in the figure.

Now, $F_1=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{a^2}$ (along $AB$ )
$F_2=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{a^2}$ (along $AD$ )
$F_3=\frac{1}{4\pi {\epsilon }_0}\frac{q_1^2}{2a^2}$
The resultant of $F_1$ and $F_2$ should be equal and opposite to $F_3$ , to keep the system in equilibrium.
Resultant of $F_1$ and $F_2$ i.e. $F_R=\sqrt{F_1^2+F_2^2}$
$=\frac{1}{4\pi }\frac{q_1{q}_1}{{ϵ}_0{a}^2}\sqrt{2}$
For equilibrium,
$\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{a^2}\sqrt{2}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1^2}{2a^2}$
$\frac{q_1}{q_2}=2\sqrt{2}$