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$\frac{q1 q2}{r^2}$

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

Answer 2

Answer:

F = 14.4 N

Explanation:

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

$F_e = \frac{kq_1q_2}{r^2}$

here we have

$q_1 = q_2 = e$

also we have

$r = 4 \times 10^{-15} m$

now we will have

$F = \frac{(9\times 10^9)(1.6 \times 10^{-19})(1.6 \times 10^{-19})}{(4 \times 10^{-15})^2}$

$F = 14.4 N$