Question

In the nucleus of an atom, two protons are separated by a distance of 10^-13 m. What is the Coloumb force between them? The charge of a proton is + 1.602 x 10^-19 C.

Answer 1

Answer:

$F=0.0231\ N$

Explanation:

Electrostatic Force

When two point charges $q_1$ and $q_2$ are placed at a distance d, they exert a force to each other whose magnitude can be computed by the Coulomb's formula

$\displaystyle F=\frac{k\ q_1\ q_2}{d^2}$

Where k is the constant of proportionality

$k=9.10^9\ Nw.m^2/c^2$

The information provided is

$d=10^{-13}\ m$

$q_1=q_2=+ 1.602 x 10^{-19}\ C$

Computing the Coloumb force between them

$\displaystyle F=\frac{9x10^9\ 1.602 x 10^{-19} \ 1.602 x 10^{-19} }{(10^{-13})^2}$

$\boxed{F=0.0231\ N}$