Calculating Electric Potential of a Point Charge Using Gauss' Law

In summary, the problem involves a point charge Q at the origin, producing an electric field of 20N/c at a radius of 2.0 meters. The goal is to find the electric potential at a spherical surface of radius 3.0 meters around the point charge, using Gauss' Law and Ohm's Law. By using Gauss' Law and plugging in the given values, the magnitude of q can be determined and then used in Ohm's Law to find the electric potential at the surface.
  • #1
ZeroCool024
8
0

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!
 
Physics news on Phys.org
  • #2
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
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
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
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?
 

Related to Calculating Electric Potential of a Point Charge Using Gauss' Law

1. What is Gauss' Law?

Gauss' Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the charge enclosed by that surface. It is named after the German mathematician and physicist, Carl Friedrich Gauss.

2. How is Gauss' Law applied to point charges?

Gauss' Law can be applied to point charges by creating a closed surface around the charge and calculating the electric flux through that surface. The electric flux is equal to the charge enclosed by the surface divided by the permittivity of free space.

3. What is the formula for Gauss' Law?

The formula for Gauss' Law is ΦE = Q/ε0, where ΦE is the electric flux, Q is the charge enclosed by the closed surface, and ε0 is the permittivity of free space.

4. How does Gauss' Law relate to Coulomb's Law?

Gauss' Law and Coulomb's Law are closely related as they both describe the electric field created by a point charge. However, Gauss' Law is more general and can be applied to any charge distribution, while Coulomb's Law only applies to point charges.

5. Can Gauss' Law be used to calculate the electric field at a point due to multiple point charges?

Yes, Gauss' Law can be applied to calculate the electric field at a point due to multiple point charges by creating a closed surface that encloses all of the charges and calculating the total electric flux through that surface. The electric field can then be determined by dividing the total flux by the permittivity of free space.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
184
  • Introductory Physics Homework Help
Replies
26
Views
725
  • Introductory Physics Homework Help
Replies
23
Views
626
  • Introductory Physics Homework Help
Replies
6
Views
876
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
875
  • Introductory Physics Homework Help
Replies
3
Views
842
  • Introductory Physics Homework Help
Replies
21
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top