The nucleus in an iron atom has a radius of about `4.0xx10^(-15)`m and contains 26 protons.
What is the magnitude of the repulsive electrostatic force between two of the protons that are separated by `4.0xx10^(-15)m`?

Explanation

To find the magnitude of the repulsive electrostatic force between two protons separated by a distance of \(4.0 \times 10^{-15}\) m, we can use Coulomb's law, which is given by the formula:

\[ F = \frac{1}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{R^2} \]

Where:
- \(F\) is the electrostatic force,
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