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 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)         0.556 m        

D)         0.350 m

E)  0.5 m

Correct Answer: C

Solution :

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