91Ó°ÊÓ

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:

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

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

The magnitude of the charge of the electron is, \(e = 1.6 \times {10^{ - 19}}\;{\rm{C}}\)

The charge of the nucleus is, \(q = 26e\).

The separation distance is, \(r = 1.5 \times {10^{ - 12}}\;{\rm{m}}\).

From expression (i), the magnitude of the electric force is,

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

Substitute the values in the above expression.

\(\begin{aligned}{l}F = \left( {9.0 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^2}} \right)\left( {\frac{{\left( {26 \times 1.6 \times {{10}^{ - 19}}\;{\rm{C}}} \right)\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C}}} \right)}}{{{{\left( {1.5 \times {{10}^{ - 12}}\;{\rm{m}}} \right)}^2}}}} \right)\\F = 2.66 \times {10^{ - 3}}\;{\rm{N}}\end{aligned}\)

Thus, the magnitude of electric force of attraction is\(2.66 \times {10^{ - 3}}\;{\rm{N}}\).