Question

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

Answer 1

Step-by-step explanation:

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