Question

4. What is the distance between two electrons, when the force between them is 0.50 N?​

Answer 1

Explanation:

$from \: coulombs \: law \: of \: electrostatics \\ F = \frac{k {q}^{2} }{ {r}^{2} } \\ 0.5 = \frac{(9 \times {10}^{9} ) \times (1.6 \times {10}^{ - 19} ) {}^{2} }{ {r}^{2} } \\ 0.5 {r}^{2} = 2.304 \times {10}^{ - 28} \\ {r}^{2} = 4.608 \times {10}^{ - 28} \\ r = \sqrt{4.608 \times {10}^{ - 28} } \\ r = 2.15 \times {10}^{ - 14} \: m$