Gauss' Law and point charges

  • #1

Homework Statement


A point charge Q is located at the origin. The point charge produces an electric field at a radius of 2.0 meters from the origin of 20N/c.The electric potential of a spherical surface of radius 3.0 meters around the point charge is, in Volts.

Homework Equations



E=kq/rs, [tex]\varphi[/tex]=Q/[tex]\epsilon[/tex]o

The Attempt at a Solution


I'm really not sure where to go with this. I'm not sure how to use gauss' law here (if i even should). Also, I'm very confused about how to convert this into electric potential. This is totally something I could clear up in like 5 min in office hours or something, but I'm deployed in Iraq right now so can't do that :frown: Thanks a ton!!!
 

Answers and Replies

  • #2
diazona
Homework Helper
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Well I have to admire your dedication :wink: You're on the right track with Gauss's law, which relates the electric field on a surface around a charge to the magnitude of the charge. How can you apply that to this problem? (Hint: what surface could you draw around the charge such that you know the electric field at every point on the surface?)
 
  • #3
Well that would be a sphere. I've considered that and trying to apply the area of a sphere to the problem in some way. One way I considered was q/4[tex]\pi[/tex]r2[tex]\varphi[/tex]. I just don't know if that's right. Or would the area go in the numerator? Assuming that is even close, in order to calculate electric potential would i multiply the electric field times the distance? Then what?
 
  • #4
ideasrule
Homework Helper
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Gauss' law is flux=q/ε0, and flux would be electric field*area for a sphere (since the electric field is the same across the area). You can then find "q".

An easier way: E=kq/r^2, right? Plug in the numbers and you can find q.

BTW, you're deployed in Iraq? Cool!
 
  • #5
So wait, to find q i don't have to use gauss' law right? I thought that it was saying the field 2 m away was 20. I could use E=q/r2. With that value of q, you can then use gauss' law to figure out the electric field over a whole sphere. So do i take that value of q and just divide it by the permitivity of free space? Thanks again a ton for the help!

P.S. yeah Iraq has been pretty interesting!
 
  • #6
ideasrule
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You can use E=kq/r^2 to determine q. Then, you can use V=kq/r to determine the electric potential at 3 m.
 
  • #7
Sorry to beat a dead horse, but even though it's looking for the electric potential at 3 m on a sphere? I just thought some crazy area of a sphere type thing might get involved, due to applying ohms law over the surface of a sphere.
 
  • #8
ideasrule
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Ohm's law works for electric circuits, and there's no circuit in this problem. If you meant Gauss' Law, that works too. It's completely equivalent to E=kq/r^2.
 
  • #9
Yeah sorry about that haha working on currents right now too. I see. So I would use the 3 m for the r term then? Just to make my stupid self sure?
 

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