question_answer

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 [Kerala PET 2002]

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 \[\left( \text{Here }k=\frac{1}{4\pi {{\varepsilon }_{0}}}\,\text{and}\,{{Q}_{1}}=5\times {{10}^{-11}}C,\,{{Q}_{2}}=-\,2.7\times {{10}^{-11}}C \right)\,\]