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Question: Two charges \(+ 4e\) and \(+ e\)are at a distance\(x\) apart. At what distance, a charge \(q\) must ...

Two charges +4e+ 4e and +e+ eare at a distancexx apart. At what distance, a charge qq must be placed from charge +e+ eso that it is in equilibrium

A

x/2x/2

B

2x/32x/3

C

x/3x/3

D

x/6x/6

Answer

x/3x/3

Explanation

Solution

For equilibrium of q

|F1| = |F2|

Which givesx2=xQ1Q2+1=x4ee+1=x3x_{2} = \frac{x}{\sqrt{\frac{Q_{1}}{Q_{2}}} + 1} = \frac{x}{\sqrt{\frac{4e}{e}} + 1} = \frac{x}{3}