Question 139 of 311

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 the $Q / q$ equals

A$-2 \sqrt{2}$B$-1$C1D$-\frac{1}{\sqrt{2}}$

Show answer and solution▾

Answer: (A) $-2 \sqrt{2}$

Three forces $F_{41}, F_{42}$ and $f_{43}$ acting on $Q$ are shown
Resultant of $F_{41}+F_{43}$
$$\begin{aligned} & =\sqrt{2} F_{\text {each }} \\ & =\sqrt{2} \frac{1}{4 \pi \varepsilon_0} \frac{Q q}{d^2} \end{aligned}$$
Resultant on $Q$ becomes zero only when ' $q$ ' charges are of negative nature.
$$\begin{aligned} & F_{4,2}=\frac{1}{4 \pi \varepsilon_0} \frac{Q \times Q}{(\sqrt{2} d)^2} \\ & \Rightarrow \sqrt{2} \frac{d Q}{d^2}=\frac{Q \times Q}{2 d^2} \\ & \Rightarrow \sqrt{2} \times q=\frac{Q \times Q}{2} \\ & \therefore q=-\frac{Q}{2 \sqrt{2}} \text { or } \frac{Q}{q}=-2 \sqrt{2} \end{aligned}$$

JEE Main Physics practice question on Electrostatics (Electric Field and Electric Field Lines), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.