Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
1915 Upvotes
Verified Answer
The correct answer is:
$-2 \sqrt{2}$
$-2 \sqrt{2}$
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.
$$
\begin{aligned}
& F_{4,2}=\frac{1}{4 \pi \varepsilon_0} \frac{Q \times Q}{(\sqrt{2} d)^2} \\
& \Rightarrow \sqrt{2} \frac{d Q}{d^2}=\frac{Q \times Q}{2 d^2} \\
& \Rightarrow \sqrt{2} \times q=\frac{Q \times Q}{2} \\
& \therefore q=-\frac{Q}{2 \sqrt{2}} \text { or } \frac{Q}{q}=-2 \sqrt{2}
\end{aligned}
$$
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.
$$
\begin{aligned}
& F_{4,2}=\frac{1}{4 \pi \varepsilon_0} \frac{Q \times Q}{(\sqrt{2} d)^2} \\
& \Rightarrow \sqrt{2} \frac{d Q}{d^2}=\frac{Q \times Q}{2 d^2} \\
& \Rightarrow \sqrt{2} \times q=\frac{Q \times Q}{2} \\
& \therefore q=-\frac{Q}{2 \sqrt{2}} \text { or } \frac{Q}{q}=-2 \sqrt{2}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.