Question

In the nucleus of an atom, two protons are separated are by a distance of 1*10^-15m. What is the magnitude of the electric force between them?
A. 115N
B. 720N
C. 142N
D. 230N

Answer 1

Answer:

D. 230 N

Explanation:

The magnitude of the electric force between the two protons is given by:

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

where

k is the Coulomb's constant

$q_1 = q_2 = 1.6\cdot 10^{-19} C$ is the charge of each proton

$r=1\cdot 10^{-15} m$ is the distance between the two protons

Substituting numbers into the formula, we find

$F=(9\cdot 10^9 N m^2 C^{-2}) \frac{(1.6\cdot 10^{-19} C)^2}{(1\cdot 10^{-15} m^2)^2}=230.4 N \sim 230 N$