The distance between charges $5 \times 10^{-11} \mathrm{C}$ and $-2.7 \times 10^{-11} C$ is 0.2 m. The distance at which a third charge should be placed in order that it will not experience any force along the line joining the two charges is
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If two opposite charges are separated by a certain distance, then for it’s equilibrium a third charge should be kept outside and near the charge which is smaller in magnitude.
Here, suppose third charge q is placed at a distance x from – 2.7 × × 10 –11 C then for it’s equilibrium |F 1 | = |F 2 |
⇒ ⇒ $\frac{kQ_{1}q}{(x+0.2)^{2}} = \frac{kQ_{2}q}{x^{2}}$ ⇒ ⇒ x = 0.556 m
$\left(\text{Here } k = \frac{1}{4 \pi \varepsilon_0} \text{ and } Q_1 = 5 \times 10^{-11} \mathrm{C}, Q_2 = -2.7 \times 10^{-11} \mathrm{C} \right)$
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