Conducting Spheres

1. Sep 23, 2006

mb85

two spheres are

Last edited: Sep 23, 2006
2. Sep 23, 2006

big man

Post a thought or two on what you are thinking and then I can post my help.

All you need to do is show that you have sort of thought about how you would do this problem.

At least tell me what you think r << d suggests to you.

3. Sep 23, 2006

mb85

well i know the charge resides on the surface of the sphere.

Eo(flux) = Qenclosed
so after taking the closed derivative i end up with

E=1/4piEo Q/r^2

which is simply E = KQ/r^2

4. Sep 23, 2006

big man

Yeah that's right for the electric field, but you are wanting to know the force of repulsion between the two charges.

Since r << d you can treat these spheres as point charges. Simply plug the numbers into Coulomb's Law using q1=q2.

Here is a hint for the second part. You know the charge of the two spheres and you know the charge of an electron (1.602*10^-19 C) so how many of those would you need for it to equal the charge on each sphere.

5. Sep 23, 2006

mb85

thats rite.

so i used F = K|q1||q2|/r^2

then i got q1 = 2.57x10^-20 C
and like u said q1 = q2

for the number of electrons...
q = ne
so i got n = 0.16

6. Sep 23, 2006

big man

That's not quite right.

From the force equation you gave you see you are multiplying q1 by q2 and since q1=q2 you essentially have q1^2.

So take the square root of the result you got and that will be your charge.
Your equation for the number is right so just fix up the charge value and you should be set.

7. Sep 23, 2006

mb85

Ok awesome. Now i got q= 1.603x10^-10 C

n = 1.00x10^9 (seems rather large tho for the number of electrons)?

thanks for all ur help. i really appreciate it.

8. Sep 23, 2006

big man

Yup that's it, good work. That's a reasonable number seeing as though the charge of an electron is extremely small.