Charge on a sphere

  • #1

Homework Statement



see attachment for pic of diagram and question

b) what is the magnitude of the charge
c) what is the number of excess or deficit electrons?

m=(2.6*10^-15)kg
Δv=265.4V
r=0.005m

Homework Equations



q=Ne
where (q=charge), (e=electron), (N=# of excess/deficit electrons)

ε=ΔV/r

FE = qε

The Attempt at a Solution



FE = Fg
qε=mg
q(ΔV/r)=mg
q=mgr/ΔV

q=(2.6*10^-15)(9.8)(0.005)/265.4
q=4.8*10^-19

The thing is, I tried checking this number and I got some weird values:

ε=kq/r^2
ε=0.0001728

FE=qε
FE=8.2944*10^-23

FE=Fg
8.2944*10^-23 = mg
8.2944*10^-23 = 2.548*10^-14

as you can see... not equal at all.
help?








oh, and assuming B) is correct, here's the answer to C):

q=Ne
N=q/e
N=(4.8*10^-19)/(1.6*10^-19)
N=3

3 deficit electrons, since charge = (+)
 

Attachments

Last edited:

Answers and Replies

  • #2
35,139
11,390
I don't see any attachment.

ε=kq/r^2
That is related to the electric field from the charged object itself, not the electric field from your setup. If you would work with units, you would see that it does not have the correct units.
 
  • #3
Ah.
What are the units on each and can you tell me of a way to check my answer? I don't see any other method (using different equations) using the set of equations I have.
 
Last edited:
  • #4
35,139
11,390
There is no alternative way to check your answer, just the formula you used to calculate the value.
Well, the nice result of "exactly" 3 electrons is an indirect way to check the result - a value of 2.5 would be weird.

What are the units on each
I think you should know/determine this yourself.
 
  • #5
ε=kq/r^2
well, i know this one is NC

the other... uhh.. I'm guessing is V/m
which I've never seen before (wasn't used in the course)
 
  • #6
35,139
11,390
V/m is the electric field strength.
If you mean N/C: Ok, then the units do not show the problem, as N/C = NV/J = V/m
 

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