Search any question & find its solution
Question:
Answered & Verified by Expert
A charge $Q$ is divided into two charges $q$ and $Q-q$. The value of $q$ such that the force between them is maximum, is
Options:
Solution:
2528 Upvotes
Verified Answer
The correct answer is:
$\frac{Q}{2}$
By Coulomb's law,
When charge $Q$ is divided into two charges $q$ and $Q-q$
$$
F=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q \cdot(Q-q)}{r^2}
$$
The value of $q$
$$
F_{\max }=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\frac{Q}{2}\left(Q-\frac{Q}{2}\right)}{r^2}
$$
(Putting $q=\frac{Q}{0}$ for the maximum force)
$$
\begin{aligned}
& =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\frac{Q}{2}\left(\frac{Q}{2}\right)}{r^2} \\
& =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\frac{Q^2}{4}}{r^2} \\
& =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q^2}{4 r^2} \\
& =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\left(\frac{Q}{2}\right)^2}{r^2}
\end{aligned}
$$
So, option (3) is correct.
When charge $Q$ is divided into two charges $q$ and $Q-q$
$$
F=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q \cdot(Q-q)}{r^2}
$$
The value of $q$
$$
F_{\max }=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\frac{Q}{2}\left(Q-\frac{Q}{2}\right)}{r^2}
$$
(Putting $q=\frac{Q}{0}$ for the maximum force)
$$
\begin{aligned}
& =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\frac{Q}{2}\left(\frac{Q}{2}\right)}{r^2} \\
& =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\frac{Q^2}{4}}{r^2} \\
& =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q^2}{4 r^2} \\
& =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\left(\frac{Q}{2}\right)^2}{r^2}
\end{aligned}
$$
So, option (3) is correct.
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.