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Physics Question on coulombs law

The electrostatic attracting force on a small sphere of charge 0.2 μC0.2\,\mu C due to another small sphere of charge −0.4 μC-0.4\,\mu C in air is 0.4 N0.4\,N. The distance between the two spheres is

A

43.2×10−6 m43.2 \times 10^{-6}\, m

B

42.4×10−3 m42.4 \times 10^{-3}\, m

C

18.1×10−3 m18.1 \times 10^{-3}\, m

D

19.2×10−6 m19.2 \times 10^{-6}\, m

Answer

42.4×10−3 m42.4 \times 10^{-3}\, m

Explanation

Solution

Here, q1=0.2μC=0.2×10−6 Cq_1 = 0.2 \mu C = 0.2 \times 10^{-6}\, C q2=−0.4 μC=−0.4×10−6 Cq_2 = -0.4 \,\mu C = -0.4 \times 10^{-6}\, C, F=−0.4 NF = -0.4\, N As F=q1 q24πε0r2F=\frac{q_{1}\,q_{2}}{4\pi\varepsilon_{0}r^{2}} ∴r2=q1 q24πε0F=0.2×10−6×0.4×10−6×9×10−90.4=1.8×10−3\therefore r^{2}=\frac{q_{1}\,q_{2}}{4\pi\varepsilon_{0}F}=\frac{0.2\times10^{-6}\times0.4\times10^{-6}\times9\times10^{-9}}{0.4} =1.8\times10^{-3} ∴r=(1.8×10−3)1/2=0.0424 m=42.4×10−3 m\therefore r=\left(1.8\times 10^{-3}\right)^{1/2}=0.0424\,m=42.4\times10^{-3}\,m