Search any question & find its solution
Question:
Answered & Verified by Expert
A charge ' $q$ ' is placed at the centre of the line joining two equal charges ' $Q$ '. The system of three charges will be in equilibrium if ' $q$ ' is equal to:
Options:
Solution:
1298 Upvotes
Verified Answer
The correct answer is:
$-Q / 4$
$$
\text { According to the question }
$$
for equilibrium, net force on charge $\quad Q=0$
$$
\therefore \frac{1}{4 \pi \epsilon_0} \frac{\mathrm{Qq}}{r^2}+\frac{1}{4 \pi \epsilon_0} \cdot \frac{\mathrm{Q}^2}{\sigma\left(\frac{r}{2}\right)^2}=0
$$
$$
\begin{aligned}
\Rightarrow & & \frac{1}{4 \pi \epsilon_0} \frac{\mathrm{Q}_2}{r^2} & =-\frac{1}{4 \pi \epsilon_0} \frac{4 \mathrm{Q} q}{r^2} \\
\Rightarrow & & Q & =-4 q \\
\Rightarrow & & q & =-\frac{Q}{4}
\end{aligned}
$$
\text { According to the question }
$$
for equilibrium, net force on charge $\quad Q=0$
$$
\therefore \frac{1}{4 \pi \epsilon_0} \frac{\mathrm{Qq}}{r^2}+\frac{1}{4 \pi \epsilon_0} \cdot \frac{\mathrm{Q}^2}{\sigma\left(\frac{r}{2}\right)^2}=0
$$
$$
\begin{aligned}
\Rightarrow & & \frac{1}{4 \pi \epsilon_0} \frac{\mathrm{Q}_2}{r^2} & =-\frac{1}{4 \pi \epsilon_0} \frac{4 \mathrm{Q} q}{r^2} \\
\Rightarrow & & Q & =-4 q \\
\Rightarrow & & q & =-\frac{Q}{4}
\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.