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Q. 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:

KEAMKEAM 2002

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.

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