Electric Force between glass ball and rubber ball

[Zokudu]

Homework Statement

What is the electric force between a glass ball with 3.5 µC of charge and a rubber ball with -5.0 µC of charge when they are separated by 5 cm?

Homework Equations

Colombs Law:
F(electric)=K((q(1)*q(2))/r^2)
K=8.99E9

The Attempt at a Solution

This seems like a simple plug in problem but I'm having a little trouble.

First i get the elementary charges of both the glass and rubber balls
so 1.6E-19*3.5=5.6E-19 for q(1)
and 1.6E-19*5=8E-19 for q(2)

So i plug in the numbers i know and get:
F(electric)=8.99E9(((5.6E-19)*(8E-19))/(.005^2))=-1.611E-22

I plug this into the program my school uses for physics homework and it tells me I am wrong.

Can you see where I am doing something wrong? please any help is appreciated.

 

[hage567]

First i get the elementary charges of both the glass and rubber balls
so 1.6E-19*3.5=5.6E-19 for q(1)
and 1.6E-19*5=8E-19 for q(2)

I don't understand why you did this. All you need to do is convert the 3.5 from microcoulombs to coulombs. Coulombs is the unit for charge.

F(electric)=8.99E9(((5.6E-19)*(8E-19))/(.005^2))=-1.611E-22

5 cm is not equal to 0.005 m. Check your conversion again.

 

[Moetasim]

Hello, it looks like you are on the right track with using Coulomb's Law to calculate the electric force between the glass and rubber balls. However, there are a few things that may be causing the discrepancy with your answer. First, the elementary charge of the glass ball should be positive since it has a charge of 3.5 µC. So, the correct value for q(1) would be 5.6E-19 C. Additionally, when using Coulomb's Law, it is important to use the distance in meters, so the separation of 5 cm should be converted to 0.05 m. Finally, make sure to include the negative sign for the charge of the rubber ball in your calculation. So, the correct equation would be:

F(electric) = 8.99E9 * ((5.6E-19 * (-5.0E-19)) / (0.05^2)) = -1.791E-22 N

I hope this helps and good luck with your physics homework!