A) 0.44 m
B) 0.65 m
C) 0.556 m
D) 0.350 m
Correct Answer: C
Solution :
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 \[\text{ }2.7\times {{10}^{11}}C\] then for it?s equilibrium \[\left| {{F}_{1}} \right|\text{ }=\text{ }\left| {{F}_{2}} \right|\] Þ \[\frac{k{{Q}_{1}}q}{{{(x+0.2)}^{2}}}=\frac{k{{Q}_{2}}q}{{{x}^{2}}}\]Þ x = 0.556 m \[\left( \text{Here }k=\frac{1}{4\pi {{\varepsilon }_{0}}}\,\text{and}\,{{Q}_{1}}=5\times {{10}^{-11}}C,\,{{Q}_{2}}=-\,2.7\times {{10}^{-11}}C \right)\,\]You need to login to perform this action.
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