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.

Step-by-step explanation:

The magnitude of charge can be calculated by the formula

`\begin{gathered} F=(kq1q2)/(r^2) \\ F=(kq^2)/(r^2) \\ q\text{ =}\sqrt{(Fr^2)/(k)} \end{gathered}`

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

`k\text{ = 9}*10^9\text{ N m}^2\text{ /C}^2`

On substituting the values, the magnitude of charge can be calculated as

`\begin{gathered} q=\text{ }\sqrt{(32.38*(0.443)^2)/(9*10^9)} \\ =\text{ 2.67}*10^(-5)\text{ C} \\ =26.7*10^(-6)\text{ C} \\ =26.7\text{ micro Coulomb} \end{gathered}`

Final Answer: The magnitude of charges is 26.7 microcoulomb.