Question

The nucleus of a helium atom contains two protons, each with a charge of 1.6 x 10-** C. If the protons are 2 x 10-4 m apart,calculate the repulsive force between them.

Answer 1

We are given the following problem:

To determine the electric force between the protons we will use Coulomb's law, that is:

$F=k_c\frac{q_1q_2}{r^2}$

Where "k" is Coulomb's constant and is equivalent to:

$k=9\times10^9\frac{Nm^2}{C^2}$

Since the charges are equal we have:

$F=k\frac{q^2}{r^2}$

Replacing the values:

$F=(9\times10^9\frac{Nm^2}{C^2})(\frac{(1.6\times10^{-19}C)^2}{(2\times10^{-15}m)^2})$

Simplifying:

$F=(9\times10^9)(\frac{1.6^2\times10^{-38}}{2^2\times10^{-30}})N$

Solving the operations:

$\begin{gathered} F=(9)(\frac{2.56}{4})(10^9)(10^{-38})(10^{30})N \\ F=57.6N \end{gathered}$

Therefore, the magnitude of the repulsive force is 57.6N.