Question

(I) What is the repulsive electrical force between two protons\({\bf{4}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 15}}}}\;{\bf{m}}\)apart from each other in an atomic nucleus?

Short answer

The repulsive electrical force between two protons is \(14.4\;{\rm{N}}\).

Step-by-step solution

Understanding the electric force

A point charge exerts an electric force on another point charge in the vicinity. It could be attractive as well as repulsive depending upon the nature of the two charges.

The expression for the electric force is given as:

\(F = k\frac{{{Q_1}{Q_2}}}{{{r^2}}}\) … (i)

Here, k is the Coulomb’s constant,\({Q_1},\;{Q_2}\)are the charges and r is the separation between the charges.

Given Data

The separation between the protons is, \(r = 4.0 \times {10^{ - 15}}\;{\rm{m}}\).

The charge of a proton is, \(q = 1.6 \times {10^{ - 19}}\;{\rm{C}}\)

Calculation of magnitude of force

From equation (i), the electric force between protons is,

\(F = k\frac{{{q^2}}}{{{r^2}}}\)

Substitute the values in the above expression.

\(\begin{aligned}{l}F = \left( {9.0 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2}} \right)\frac{{{{\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C}}} \right)}^2}}}{{{{\left( {4.0 \times {{10}^{ - 15}}\;{\rm{m}}} \right)}^2}}}\\F = 14.4\;{\rm{N}}\end{aligned}\)

Thus, the repulsive electrical force between two protons is\(14.4\;{\rm{N}}\).