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Question: The electrostatic force on a small sphere of charge \(0.2\mu C\)due to another small sphere of charg...

The electrostatic force on a small sphere of charge 0.2μC0.2\mu Cdue to another small sphere of charge −0.4μC- 0.4\mu Cin air is 0.40.4N. The distance between the two spheres is.

A

4.2×10−6m4.2 \times 10^{- 6}m

B

4.2×10−3m4.2 \times 10^{- 3}m

C

1.8×10−3m1.8 \times 10^{- 3}m

D

1.8×10−6m1.8 \times 10^{- 6}m

Answer

4.2×10−3m4.2 \times 10^{- 3}m

Explanation

Solution

: Here, q1=0.2μC=0.2×10−6Cq_{1} = 0.2\mu C = 0.2 \times 10^{- 6}C

q2=−0.4μC=0.4×10−6Cq_{2} = - 0.4\mu C = 0.4 \times 10^{- 6}C

F = 0.4 N

As F=q1q24πε0r2F = \frac{q_{1}q_{2}}{4\pi\varepsilon_{0}r^{2}}

∴r2=q1q24πε0F=0.2×10−6×0.4×10−6×9×1090.4\therefore r^{2} = \frac{q_{1}q_{2}}{4\pi\varepsilon_{0}F} = \frac{0.2 \times 10^{- 6} \times 0.4 \times 10^{- 6} \times 9 \times 10^{9}}{0.4}

=1.8×10−3=0.18×10−4= 1.8 \times 10^{- 3} = 0.18 \times 10^{- 4}

∴r=(0.18×10−4)1/2=0.42×10−2m=4.2×10−3m\therefore r = (0.18 \times 10^{- 4})^{1/2} = 0.42 \times 10^{- 2}m = 4.2 \times 10^{- 3}m