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The distance between charges $5 \times {10^{ - 11}}\,C$ 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......$m$
$0.44$
$0.65$
$0.556$
$0.350$
Solution
(c) 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 -$2.7 \times 10^{-11}C$ then for it’s equilibrium $|F_1| = |F_2|$
$==>$ $\frac{{k{Q_1}q}}{{{{(x + 0.2)}^2}}} = \frac{{k{Q_2}q}}{{{x^2}}}$ $==>$ $x = 0.556\, m$
$\left( {{\rm{Here }}\,k = \frac{1}{{4\pi {\varepsilon _0}}}\,{\rm{and}}\,{Q_1} = 5 \times {{10}^{ – 11}}\,C,\,{Q_2} = – \,2.7 \times {{10}^{ – 11}}\,C} \right)\,$
