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Q.

If the electrostatic attracting force on a small sphere of charge 0.2 μC due to another small sphere of charge -0.4 μC in air is 0.4 N. The distance between the two spheres will be

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a

42.4 x 10-3 m

b

 19.2 x 10-6 m 

c

43.2 x 10-6 m

d

 18.1 x 10-3 m 

answer is B.

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Detailed Solution

Here, q1=0.2 μC=0.2×10-6C

q2=-0.4 μC=-0.4×10-6 C, F=-0.4 N

As F=q1q24πε0F=0.2×10-6×0.4×10-6×9×1090.4

⇒ r2=1.8×10-3

Therefore r =(1.8×10-3)1/2=0.0424 m=42.4×10-3 m

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