# Calculate the number of excess electrons

## 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.

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?

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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.

would this equation be all right?

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

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