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(q_1q_2)/(r^2)`

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

`k=9*10^9(Nm^2)/(C^2)`

Since the charges are equal we have:

`F=k(q^2)/(r^2)`

Replacing the values:

`F=(9*10^9(Nm^2)/(C^2))(((1.6*10^(-19)C)^2)/((2*10^(-15)m)^2))`

Simplifying:

`F=(9*10^9)((1.6^2*10^(-38))/(2^2*10^(-30)))N`

Solving the operations:

`\begin{gathered} F=(9)((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.