Two charged particles are placed at a distance of 1.0 cm apart. What is the minimum possible magnitude of electrostatic force acting on each charge.

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25 x10-²⁴N

2.3 x 10^-24N

23 x10^-24N

2.3 x10^-24 dyne

answer is D.

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Detailed Solution

$\begin{array}{l} F_{min} = \frac{1}{4 {πε}_{0}} \cdot \frac{e \times e}{r^{2}} = \frac{9 \times 10^{9} \times 1 .6 \times 10^{- 19} \times 1 .6 \times 10^{- 19}}{{(1 \times 10^{- 2})}^{2}} \\ = 2 .3 \times 10^{- 24} N \end{array}$
(min. charge = one electron charge)

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