question_answer

The distance between charges\[5\times {{10}^{-11}}C\]and\[-2.7\times {{10}^{-11}}C\] is 0.2m. The distance at which is third charge should be placed from 4e in order that it will not experience any force along the line joining the two charges is

A)  0.44 m          

B)  0.65 m

C)  0.556m         

D)  0.350m

Correct Answer: C

Solution :

From the question\[{{F}_{1}}={{F}_{2}}\] \[5\times {{10}^{-11}}C-2.7\times {{10}^{-11}}C\] \[\Rightarrow \] \[\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{5\times {{10}^{-11}}\times q}{{{(0.2+x)}^{2}}}\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{2.7\times {{10}^{-11}}\times q}{{{x}^{2}}}\] \[\Rightarrow \] \[x=0.556m\] from\[IInd\]charge.