Electrostatics Question 346

Question: Two charges $ +4e $ and $ +e $ are at a distance$ x $ apart. At what distance, a charge $ q $ must be placed from charge $ +e $ so that it is in equilibrium

Options:

A) $ x/2 $

B) $ 2x/3 $

C) $ x/3 $

D) $ x/6 $

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Answer:

Correct Answer: C

Solution:

For equilibrium of q |F1| = |F2| Which gives $ x _{2}=\frac{x}{\sqrt{\frac{Q _{1}}{Q _{2}}}+1}=\frac{x}{\sqrt{\frac{4e}{e}}+1}=\frac{x}{3} $



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