Calculate the number of excess electrons

[Mitchtwitchita]

Homework Statement

Two negatively charged objects repel each other with a measured force of 6.3 N when they are 0.5 cm apart. If the excess charge on one of the objects is caused by 8.3 x 10^22 extra electrons, use Coulomb's Law to calculate the number of excess electrons on the second object.

Homework Equations

F=kq1q2/r^2

The Attempt at a Solution

Since 1e=1.602 x 10^-19C
8.3 X 10^22e = (1.602 x 10^-19 C)(8.3 x 10^22)
=1.3 x 10^4 C

Therefore, 6.3N = (9 x 10^9 N*m^2/C^2)1.3 x 10^4 C)(q2)/(0.005m)^2
=1.3 x 10^-18 C

Since 1C = 6.242 x 10^18e
1.3 x 10^-18C = (1.3 x 10^-18)(6.242 x 10^18)e
=8.1e ?


I think I might be over complicating this one. Can anybody help me out? Please?

 

[Mentz114]

Hi, i don't understand what you've done.

6.3N = (9 x 10^9 N*m^2/C^2)1.3 x 10^4 C)(q2)/(0.005m)^2
=1.3 x 10^-18 C

What is this ? 6.3N=1.3x10^-8 C ?

If you write a=b then a should equal b. You are confused.
Here's a tip - don't put in the numbers until you have worked out an algebraic formula for what you want. You want q1, and then to divide by the electronic charge.

 

[Mitchtwitchita]

would this equation be all right?

q1=F(r)^2/k(q2)

 

[Mentz114]

Yes, good, but better written like this ( because F(r) looks like a function)

q1 = F*r^2/(k*q2)

It's also clearer if you show the multiplication sign. Now plug in the numbers and then divide by the electronic charge.

 

[Mitchtwitchita]

Thanks Mentz! I was over complicating it. Smooth sailing now though!