Question

Two equal, but oppositely charged particles are attracted to each other electrically. The size of the force of attraction is 48.1 N when they are separated by 60.9 cm. What is the magnitude of the charges in microCoulombs?

Answer 1

Given:

The force of attraction is F = 48.1 N

The separation between the charges is

$\begin{gathered} r=\text{ 60.9 cm} \\ =\text{ 60.9}\times10^{-2}\text{ m} \end{gathered}$

Also, the magnitude of charge q1 = q2 = q.

To find the magnitude of charge.

Explanation:

The magnitude of charge can be calculated by the formula

$\begin{gathered} F=\frac{k(2q)}{r^2} \\ q=\frac{Fr^2}{2k} \end{gathered}$

Here, k is the Coulomb's constant whose value is

$k\text{ = 9}\times10^9\text{ N m}^2\text{ /C}^2$

On substituting the values, the magnitude of charge will be

$\begin{gathered} q=\frac{48.1\times(60.9^\times10^{-2})^2}{2\times9\times10^{^9}} \\ =9.91\times10^{-10}\text{ C} \\ =9.91\times10^{-4}\text{ }\mu C \end{gathered}$

Thus, the magnitude of each charge is 9.91 x 10^(-4) micro Coulombs.