Chapter 16, Problem 3P

To determine

The magnitude of the electric force of attraction.

Answer to Problem 3P

The magnitude of the electric force of attraction is $2.65\times {10}^{-3}\text{N}$ .

Explanation of Solution

Given information:

Magnitude of charge on an iron nucleus, $Q_1=26e$

Magnitude of charge on a proton, $Q_2=e$

Distance between charges, $r=1.5\times {\text{10}}^{-12}\text{m}$

Formula used:

Write the expression to calculate the magnitude of force acting between two charges using Coulomb’s law.

  $F=k\frac{Q_1{Q}_2}{r^2}$........... (1)

Where,

The force acting between two charges is $F$ .

The Coulomb’s constant is $k$ and its value is $8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^2$ .

Calculation:

Since, the charge on the innermost electron is equal to the charge on the proton.

Substitute the values in equation (1).

  $\begin{array}{c}F=\left(8.988\times {10}^9\right)\frac{\left(26e\right)\left(e\right)}{{\left(1.5\times {\text{10}}^{-12}\right)}^2}\\ =8.988\times {10}^9\times 1.155\times {10}^{25}\times {\left(1.6\times {10}^{-19}\right)}^2\\ =2.65\times {10}^{-3}\text{N}\end{array}$

Therefore, themagnitude of the electric force of attraction is $2.65\times {10}^{-3}\text{N}$ .