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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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$25 \times 10^{- 24} N$

$23 \times 10^{- 24} N$

$2.3 \times 10^{- 24} d y n e$

$2.3 \times 10^{- 24} N$

answer is D.

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Detailed Solution

$F = \frac{9 \times 10^{9} \times e \times e}{r^{2}}$

$= 2.3 \times 10^{- 24} N$

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