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Q. Two charges $ 4q $ and $ q $ are placed at distance $ l $ . In the middle adjoining line a charge $ Q $ is kept. If resultant force on $ q $ will be zero, then $ Q $ will be

Rajasthan PETRajasthan PET 2005

Solution:

Here $ {{F}_{AB}}+{{F}_{CB}}=0 $
$ \frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{4{{q}^{2}}}{{{l}^{2}}}+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{Qq}{{{l}^{2}}/4}=0 $
$ \Rightarrow $ $ 4Qq=-4{{q}^{2}} $
$ \Rightarrow $ $ Q=-q $

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