Electrostatics 1 Question 10
10. 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) $+\frac{Q}{2}$
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Answer:
Correct Answer: 10. (d)
Solution:
- Since, $q$ is at the centre of two charges $Q$ and $Q$, net force on it is zero, whatever the magnitude and sign of charge
$$ \dot{Q} \quad \dot{q} \quad \dot{Q} $$
on it. For the equilibrium of $Q, q$ should be negative because 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
$$ q=\frac{Q}{4} $$
