Electrostatics Ques 9
- A charge $q$ is placed at the centre of the line joining two equal charges $Q$. The system of the three charges will be in equilibrium if $q$ is equal to
(1987, 2M)
(a) $-\frac{Q}{2}$
(b) $-\frac{Q}{4}$
(c) $+\frac{Q}{4}$
(d) $+\frac{Q}{2}$
Show Answer
Answer:
Correct Answer: 9.(b)
Solution:
Formula:
Coulomb Force Between Two Point Charges:
Since, $q$ is at the centre of two charges $Q$ and $Q$, net force on it is zero, whatever the magnitude and sign of $q$
on it. For the equilibrium of $Q, q$ should be negative because the other charge $Q$ will repel it, so $q$ should attract it. Simultaneously these attractions and repulsions should be equal.
or
$$ \frac{1}{4 \pi \varepsilon_{0}} \frac{Q Q}{r^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q q}{(r / 2)^{2}} $$
or with sign
$$ q=\frac{Q}{4} $$
BETA
