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Q.

The smallest electric force between two charges placed at a distance of 2m is (x+0.76)1029 N. Find value of x14πε0=9×109 SI unit 

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answer is 5.

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Detailed Solution

F=14π0q1q2r2

For F to be smallest

q1,q2 must be smallest

i.e  q1=q2=1.6×109C

Fmin=9×109×1.6×1.6×10384=9×0.64×1029

F=5.76×1029N

F=(5+0.76)1029Nx=5

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