How close must two electrons be if the magnitude of the electric force between them is equal to the weight of either at the Earth’s surface?

According to Coulomb’s law, the magnitude of electric force (F) that one small object (having charge\({{\bf{Q}}_{\bf{1}}}\)) exerts on another small object (having charge\({{\bf{Q}}_{\bf{2}}}\)) is directly proportional to the product of the charges on both the objects and inversely proportional to the square of the distance (r) between them.

The expression for electric force is given as:

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

Here, k is electrostatic force constant whose value is \(9.0 \times {10^9}\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}\)

Mass of each electron is, \(m = 9.11 \times {10^{ - 31}}\;{\rm{kg}}\)

The magnitude of the charge on each electron is, \(q = 1.602 \times {10^{ - 19}}\;{\rm{C}}\)

Let the separation between the two charges be r.

The magnitude of electric force between the two electrons is given as:

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

The weight of each electron at the Earth’s surface is:

\(W = mg\)

Here, g is the acceleration due to gravity.

According to the question, the magnitude of the electric force between two electrons is equal to the electron's weight. Then,

\(\begin{aligned}{c}F = W\\k\frac{{{q^2}}}{{{r^2}}} = mg\\r = q\sqrt {\frac{k}{{mg}}} \end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}{c}r = 1.602 \times {10^{ - 19}}\;{\rm{C}}\sqrt {\frac{{9.0 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}}}{{\left( {9.11 \times {{10}^{ - 31}}\;{\rm{kg}}} \right) \times \left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)}}} \\ = 1.602 \times {10^{ - 19}}\;{\rm{C}} \times 3.175 \times {10^{19}}\;{\rm{m/C}}\\ = 5.08\;{\rm{m}}\end{aligned}\)

Thus, two electrons must be at a distance of 5.08 m from each other.