Question

Two equal, but oppositely charged particles are attracted to each other electrically. The size of the force of attraction is 32.38 N when they are separated by 44.3 cm. What is the magnitude of the charges in microCoulombs ?

Answer 1

Given:

There are two equal and opposite charged paticles.

The force of attraction is F = 32.38 N

The distance between the charged particles is r = 44.3 cm =0.443 m

Required: Magnitude of each charge in micro coulombs.

Explanation:

The magnitude of charge can be calculated by the formula

$\begin{gathered} F=\frac{kq1q2}{r^2} \\ F=\frac{kq^2}{r^2} \\ q\text{ =}\sqrt{\frac{Fr^2}{k}} \end{gathered}$

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 can be calculated as

$\begin{gathered} q=\text{ }\sqrt{\frac{32.38\times(0.443)^2}{9\times10^9}} \\ =\text{ 2.67}\times10^{-5}\text{ C} \\ =26.7\times10^{-6}\text{ C} \\ =26.7\text{ micro Coulomb} \end{gathered}$

Final Answer: The magnitude of charges is 26.7 microcoulomb.