What do we mean

1. Sep 18, 2007

pardesi

how does coloumb's law accomadate this
the field on a say a uniformly charged shell.any point on the shell is charged so does that point contribute to the field there itself ye s the question will look stupid but does it?
further does gauss's law blow up when i take the charged sphere itself to be my gaussian surface

2. Sep 18, 2007

pardesi

also what is the field on a metallic charged sphere

3. Sep 18, 2007

Loren Booda

Your Gaussian sphere, encompassing the surface charge Q, would lie on a spherical surface just above the charged sphere. In general, the charge would distribute itself over a metallic surface so that there would exist an averaged surface density, Q/4(pi)R2, which can be integrated to yield Gauss's law.

4. Sep 18, 2007

pardesi

what if my gaussian surface were the sphere(shell) itself

5. Sep 18, 2007

Loren Booda

What if the shell were defined by the charges (electrons)? Then it becomes a quantum problem, out of the reign of Gauss's law. Look up solid state physics and energy bands.

6. Sep 18, 2007

StatusX

Please don't look up energy bands. The point is that the model of a sphere with an infinitely abrupt edge is an idealization, and a better approximation would have the charge density highest near the surface, and falling off continuously (though very quickly) as you move away. Basically, you just can't apply Gauss' law for the surface you chose around a perfect sphere, but there's no such problem for any realistic example.

7. Sep 19, 2007

pardesi

well my actual problem is how do we define the net electric field at a point where the charges are present say inside a solid sphere

8. Sep 19, 2007

StatusX

If the charge at the point is a charge density, so that the smaller a region you look at near the point, the less charge is in it, then there's no problem: just look at a small enough region so that the charge in that region can be ignored. If it's a point charge, then this won't help, and in fact there's no well defined way to define the field right at such a point.

9. Sep 19, 2007

pardesi

well what about conductors we know that for sufficiently close toa surface the field at that point(or near that) due to that charge is half of the total field above it so that surely can't be ignored

10. Sep 19, 2007

StatusX

When I say charge density, I mean a volume density. Surface/line/point charges are all singularities, associated with a discontinuity of the field, and it is usually impossible to consistenly assign a value of the electric field right at such a singularity. In any case, the value of a field at a single point, or even along a line or surface, is meaningless, because a charge will only experience the field there for an infinitessimally short time (unless its confined to the line or plane, in which case you really have a <3 dimensional problem), so it will not have any significant effect on its dynamics. What's important is the value of a field in a neighborhood of each point.

Last edited: Sep 19, 2007
11. Sep 19, 2007

pardesi

thanks for that also i was going through two proofs in which the
1.gaussian surface was a pillbox(match-box) through a surface charge distribution was used how do i define the flux through these
as an example take the standard problem of an infinite sheet of charge ho is the flux through the pillbox gaussian surface define there
2.calculating the potential of a shell inside with infinity as refrence point then surely u have to pass through the outer cover while calculating the line integral how do u hen define the line integral

12. Sep 19, 2007

StatusX

Before applying Gauss' law it's usually helpful to take note of the symmetries of the problem. In the case of an infinite sheet of charge, there is a symmetry of rotation about the z axis (where the sheet is in the xy plane). This means the electric field can only have a z component. There is also a symmetry of translation in the xy directions, so the magnitude of the field can only depend on z (let me know if you're not familiar with these kinds of arguments). Finally, there is a symmetry of reflection through the plane, so the field at z is the reflection of the field at -z.

So we take the pillbox to straddle the sheet, and so that its faces are parallel to the xy,yz, and xz planes. Then the vertical faces have no flux, since the field is perpendicular to them, and the flux of the two horizontal faces is just equal to their area times the field at a point on the face, since the field is constant across the face. This allows you to solve for the field.

For the shell, you do have to pass through the shell when calculating the potential, but because the field is doesn't blow up, the contribution to the line integral from just around the shell is negligible, and doesn't affect the calculation. This is just an example of the integral of a non-continuous function (such as f(x)=-1, x<0, f(x)=1, x>0) being a continuous function (in this case, |x|, which is true regardless of what you set f(0) to be).

Last edited: Sep 19, 2007
13. Sep 20, 2007

pardesi

thanks for the second one.
yes i am accustomed to the arguemnet u presented in the first well my corncern was not the proof but how do u define field at the point where the gaussian surface cuts the plane sheet

14. Sep 20, 2007

StatusX

It doesn't matter. Because the intersection is a line, its contribution to the integral is zero, whatever the value of the field there is.

15. Sep 26, 2007

pardesi

how do we define electric field inside a sphere?although the charge may be less but at that point dietance is also infetismal?