Calculating Electric Potential of a Point Charge Using Gauss' Law

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Homework Help Overview

The discussion revolves around calculating the electric potential of a point charge using Gauss' Law. The original poster presents a scenario involving a point charge at the origin and seeks to determine the electric potential at a spherical surface surrounding the charge.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the application of Gauss' Law and its relationship to electric fields and potentials. There are attempts to connect the electric field at a specific radius to the charge and potential, with questions about the correct formulas and methods for calculating these values.

Discussion Status

The discussion is active, with participants providing guidance on using Gauss' Law and the relationship between electric field and potential. Multiple interpretations of how to approach the problem are being explored, particularly regarding the use of different formulas and the relevance of spherical geometry.

Contextual Notes

Participants note constraints such as the original poster's inability to seek in-person help due to deployment. There are also discussions about the appropriateness of applying concepts from different areas of physics, such as Ohm's Law, to this problem.

ZeroCool024
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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!
 
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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?)
 
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?
 
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!
 
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!
 
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.
 
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.
 
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.
 
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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