The electrostatic force on a small sphere of charge 0.4 pC due to another small sphere of charge – 0.8 μC in air is 0.2

Question

The electrostatic force on a small sphere of charge 0.4 pC 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?

Answer 1

q₁ = 0.4 μC = 0.4 x 10^-6 C, q₂ = – 0.8μC = – 0 .8 x 10^-6 C, F = 0.2 N

(a) F = 9 x 10⁹
r² = 9 x 10⁹ x \(\frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{~F}}\)
= \(\frac{9 \times 10^{9} \times 0.4 \times 10^{-6} \times 0.8 \times 10^{-6}}{0.2}\)
= 14.4 x 10^-3
= 144 x 10^-4m²
r = 12 x 10^-2m

(b) 0.2 N, attractive.