Gauss's law doesnt work here?

1. Aug 6, 2008

atavistic

So I was doing this one problem in which there is a sphere of radius r and charge density row and there is a spherical cavity inside it whose center is at a distance r2 from the center of the bigger sphere.

Now the question was what is the E field inside the cavity.

My first and long thought was that there is no charge inside this cavity, so I will draw a gaussian surface inside this cavity and notice that flux is 0 and say E is 0. After getting the wrong answer, and discussing with some friends , they said to consider superposition of a sphere of charge density -row and then find E and I got the right answer but then why does Gauss law fail in giving me the right result?

This might be very elementary but I have just started.

2. Aug 6, 2008

Staff: Mentor

What Gauss's law will tell you is that since there is no charge within the cavity, the flux must be zero. But that does not mean that the field is zero!
Gauss's law is only useful for finding E when there is some symmetry that you can take advantage of, which is not the case here since the cavity is off-center and breaks the symmetry.

3. Aug 20, 2008

mrivano

how do you derive equation of E on this question?
i was thinkin of using superposition
E=Ebigsphere-Elittlesphere

but as the cavity has no charge, then Elittlesphere would be zero which result in E=Ebigsphere.

how to solve this? im completely lost!! thanks

4. Aug 20, 2008

atyy

That's curious. Is Gauss's law really giving you the wrong answer?

5. Aug 20, 2008

Gear300

and there you have it

6. Aug 20, 2008

Defennder

Yes, I'm puzzled too as to why the E-field in the cavity is non-zero.

Now consider any point in the cavity (denote this by A) where we assume the potential difference with the cavity surface is non-zero. A closed equipotential surface, which A lies on, could be drawn within the cavity. Now consider a Gaussian surface which encloses this equipotential surface, but which itself is enclosed by the cavity. If the E-field in the cavity is non-zero, then we should expect the total outward (or inward) electric flux (from the inside equipotential surface to the cavity surface or in the opposite direction) through this surface to be non-zero, which by Gauss law implies that there is a charge in the cavity, which is certainly not true.

What is your teacher's explanation for why the E-field is non-zero?

7. Aug 20, 2008

Defennder

I'm not convinced as to why that is the explanation. Sure the symmetry isn't perfect, but we can conceive of a general scenario for which Gauss law is violated if we assumed non-zero E-field.

8. Aug 20, 2008

atyy

Yup, the lack of symmetry is the right answer. I didn't read the question correctly, thought the two spheres were concentric, and that you were supposed to find the field at r/2. I assume the inner sphere is a conductor?

Gauss's law still works - after you solve for the E field the hard way, calculate the flux on your Gaussian surface, and you'll find that it is zero. In the case of a field with no spherical symmetry, it must be that some flux is pointing into the Gaussian surface, and some flux is pointing out of it, and those all cancel. Gauss's law works, but it just happens to be not very useful in this case. It's useful if you know the direction of the E field in advance from symmetry, then you can make a clever choice for the shape of the Gaussian surface to give you the magnitude of the E field.

9. Aug 20, 2008

Staff: Mentor

Take any Gaussian surface within the cavity. Just because the E-field is non-zero does not imply that the flux through a closed Gaussian surface is non-zero. The E-field is non-zero, the flux is zero, the charge within the cavity is zero: All consistent with Gauss's law. (And the Gaussian surface won't be an equipotential--don't know where that came from.)

10. Aug 20, 2008

atyy

OK, I'm confused about a few things. Is the outer sphere a conductor? Is the inner sphere a conductor? Seems we need to know whether the charges can redistribute themselves. For example, the total charge on the inner conductor might be zero, but if the charges can move, then we may end up with a situation where the negative charge is gathered on one side, and the positive charge on another.

Actually, since you told us that it's equivalent to pretending that the inner sphere has charge density of -rho, that's equivalent to telling us which spheres are conducting ...

11. Aug 20, 2008

atyy

OK, I get it I think. It's not a sphere of charge (ie. with charge on its surface only). It's a solid ball of charge with a hole that's off center. All the charge is fixed in place, so these are not conductors. That makes sense, since volume charge density is usually denoted rho, but surface charge density is usually denoted sigma.

12. Aug 20, 2008

Staff: Mentor

Exactly. This is not a conductor.

13. Aug 20, 2008

Staff: Mentor

To understand the difference between the E-field and the flux through a Gaussian surface, consider Gauss' law applied to the space between the plates of a capacitor. There is a uniform E-field in this region, but no charges. The E-field is non-zero, but the flux is zero because any E-field line that enters the surface must also leave the surface on the other side.

Whenever I get confused about flux v. field that is the example I think about.

14. Aug 20, 2008

Defennder

You're right. I mistakenly assumed it was a conductor.

15. Aug 20, 2008

Staff: Mentor

Or consider a closed Gaussian surface with a point charge outside but no charge inside. Any field line (going radially outward from the charge) that enters the surface has to come out somewhere, so the net flux through the surface is zero.

16. Aug 20, 2008

atyy

OK, what about an outer nonconducting sphere with uniform surface charge density sigma. Put a conducting sphere inside it that's off center. The inner conducting sphere has no net charge.

Will the E field inside the outer sphere be zero?
(Edit: I think it will be, because if the outer sphere were a conductor, that would be a Faraday cage, and there would be no E field inside. If it were a conductor, the charge would be uniformly distributed. So the uniform charge distribution on the non-conducting sphere mimics that conducting sphere, so the E field will be zero)

If it is, are we justified in using Gauss's law to do so, or does that just accidentally give the right answer, since we don't have spherical symmetry?
(Edit: For example, how do we know that there isn't some separation of charge on the inner sphere, since part of it is nearer the charged outer wall than another part, since it is off center?)

Last edited: Aug 20, 2008