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NCERT Class XII Chapter
Electric Charges and Fields
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Question : 2 of 34
Marks: +1, -0
The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge –0.8 μC in air is 0.2 N.
(a) What is the distance between the two spheres?
(b) What is the force on the second sphere due to the first?
Solution:  
Given, F12F_{12} = 0.2 N , q1q_1 = 0.4 × 10−610^{-6} C , q2q_2 = - 0.8 × 10−610^{-6} C
As, F12F_{12} = 14πϵ0q1q2r2\frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} or r2r^2 = 14πϵ0q1q2F12\frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{F_{12}} = 9 × 10910^9 × 0.410−6×0.8×10−60.2\frac{0.410^{-6} \times 0.8 \times 10^{-6}}{0.2}
or r2r^2 = 14.4 × 10−310^{-3} = 1.44 × 10−210^{-2}
or r = 1.2 × 10−110^{-1} m = 12 cm
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