Three charges $q, Q$ and $+4 q$ are placed in a straight line of length $d$ at points at distances $0,…
- $-2 q$
- $\frac{-q}{2}$
- $-q$
- $\frac{-3}{2} q$
Solution
To make net force on $q$ is zero, charge $Q$ should be negative.
$\begin{aligned}
& \therefore \frac{k q Q}{\left(\frac{d}{2}\right)^2}+\frac{k q(4 q)}{d^2}=0 \\
& \Rightarrow Q=-q
\end{aligned}$Asked in: MHT CET 2022 (11 Aug Shift 1)