Question

What is the repulsive electrical force between two protons 4.0×10−15m apart from each other in an atomic nucleus?

Answer 1

F = K
`(q1 q2)/(r^2)`

F = 1/(4 * 3.14 * 8.85 * 10^-12) [ (1.6x 10^-19)^2/(4 x 10 ^-15)^2 ] = 14.394 N

Answer 2

F = 14.4 N

Step-by-step explanation:

As we know that the force between two charge particles is given by coulombs law as

`F_e = (kq_1q_2)/(r^2)`

here we have

`q_1 = q_2 = e`

also we have

`r = 4 * 10^(-15) m`

now we will have

`F = ((9* 10^9)(1.6 * 10^(-19))(1.6 * 10^(-19)))/((4 * 10^(-15))^2)`

`F = 14.4 N`