Question

Two equal, but oppositely charged particles are attracted to each other electrically. The size of the force of attraction is 56.46 N when they are separated by 50.91 cm. What is the magnitude of the charges in microCoulombs?

Answer 1

Answer:

q = 40.3 microCoulombs

Step-by-step explanation:

The force of attraction between two charges is given by the formula:

`F=(kq^2)/(r^2)`

The force of attraction. F = 56.46 N

The separation, r = 50.91 cm

r = 50.91/100

r = 0.5091 m

The electrostatic constant is:

`k=9*10^9Nm^2C^(-2)`

Solve for the magnitude of charge q

`\begin{gathered} F=(kq^2)/(r^2) \\ \\ 56.46=(9*10^9* q^2)/(0.5091^2) \\ \\ 56.46*0.5091^2=9*10^9* q^2 \\ \\ q^2=(56.46*0.5091^2)/(9*10^9) \\ \\ q^2=1.63*10^(-9) \\ \\ q=\sqrt{1.63*10^(-9)} \\ \\ q=4.03*10^(-5)C \\ \\ q=40.3\mu C \end{gathered}`