Electric Force between glass ball and rubber ball

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SUMMARY

The electric force between a glass ball with a charge of 3.5 µC and a rubber ball with a charge of -5.0 µC, separated by 5 cm, can be calculated using Coulomb's Law. The correct formula is F(electric) = K((q(1)*q(2))/r^2), where K = 8.99E9. The charges should be converted to coulombs before calculation, resulting in q(1) = 3.5E-6 C and q(2) = -5.0E-6 C. The distance must also be converted to meters, making it 0.05 m, leading to a corrected calculation of the electric force.

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  • Unit conversion from centimeters to meters
  • Basic algebra for solving equations
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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.
 
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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.
 

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