Electric field outside conductor surface

[Elmer Correa]

I'm confused as to how this expression E= (δ/εo) can't be used to calculate the electric field of a perfectly flat part of a surface even farther from just above the surface.If you just extend the same cylindrical Gaussian surface used for this proof, wouldn't the field stay the same no matter how far out you go? The charge enclosed on the surface of the conductor wouldn't change and neither would the area of the circular end of the cylinder sticking out of the conductor since it sticks out of a flat surface. Obviously this simply cannot be so as the field should grow weaker with distance from its charged source, so I wonder what about this proof I'm misunderstanding. On a related note, what about the derivation for E=(δ/2εo) for an infinitely charged plane takes into account its infinite size, as it seems to me that it could just as easily be used to calculate the electric field for a finite sheet of charge?

 

[mishima]

I think this is a common confusion about conducting surfaces (which have some volume) vs sheets of charge. In the former (no factor of 1/2) you have a slab of conducting material with 2 sheets of charge on opposite sides. For a sheet of charge (equation with factor 1/2) you have 1 sheet of charge.

The object with double the sheets of charge is exactly double the field of a single sheet.

 

[Elmer Correa]

Yes, I see this now, but can you explain why the electric field above a flare surface of a charged conductor wouldn't be constant using a Guassian cylindrical surface?

 

[mishima]

Sorry, maybe I misunderstood your original question.

Field magnitude can be interpreted as density of the field lines, right? A cylindrical Gaussian surface gets flux through its circular end caps only. For both types of surface, we're usually talking about a field that is perpendicular to the surface.

Compare a very small tube with one line of flux shooting through perpendicularly, with a very long tube with one line of flux shooting through perpendicularly. In other words the density of the field lines is the same in both situations at the end caps.

 

[Elmer Correa]

Sorry, maybe I misunderstood your original question.

Field magnitude can be interpreted as density of the field lines, right? A cylindrical Gaussian surface gets flux through its circular end caps only. For both types of surface, we're usually talking about a field that is perpendicular to the surface.

Compare a very small tube with one line of flux shooting through perpendicularly, with a very long tube with one line of flux shooting through perpendicularly. In other words the density of the field lines is the same in both situations at the end caps.

I suppose I may not have explained myself well, but going off of your last paragraph where the size of the tube does not matter, how could we not apply this to a flat surface on a charged conductor? Wouldn't the electric field perpendicular to this flat part at any distance be the same, regardless of whether it's infinite or not?

 

[mishima]

Basically in the finite case (like a charged disk, or L*W square), the field gets some curvature. In the inifinite case, its just perpendicular everywhere.

If you grasp a bundle of long grass, near your hand the strands seem to be parallel to each other (resembling the infinite case) but eventually the strands diverge.

In that picture, suppose my long tube is centered on the right very far from the disk. It will have different flux lines through the circular end caps of the cylindrical Gaussian surface out there because the curvature is more noticeable with distance. Nearer to the disk though, the same lines are essentially all parallel (all perpendicular to the surface).

 

[NFuller]

When you use Gauss's Law, you are integrating $\mathbf{E}\cdot d\mathbf{a}$ where $d\mathbf{a}$ is an infinitesimal area vector normal to the Gaussian surface. For some problems we can use symmetry arguments to simplify this dot product and not really do any integration. For an infinite sheet of charge, the surface charge looks the same in all directions, so the field $\mathbf{E}$ must also look the same in all directions. This implies that $\mathbf{E}$ is pointing straight up from the surface and is uniform. In this case, if you use a cylindrical Gaussian surface, the integral simplifies to
$$\oint_{s}\mathbf{E}\cdot d\mathbf{a}=\int E_{z}da=E_{z}\int da=\frac{Q}{\epsilon_{0}}$$
The last surface integral is just the surface area of both end caps $2\pi r^{2}$ and the inclosed charge is simply $Q=\sigma\pi r^{2}$. This leads us to
$$E_{z}=\frac{\sigma}{2\epsilon_{0}}$$

For a finite surface however, this symmetry argument does not hold. The charge distribution does change, specifically at the edge of the surface where it goes to zero. So we cannot be sure of the direction of $\mathbf{E}$ and cannot make any simplifications to the integral.

 

[Elmer Correa]

When you use Gauss's Law, you are integrating $\mathbf{E}\cdot d\mathbf{a}$ where $d\mathbf{a}$ is an infinitesimal area vector normal to the Gaussian surface. For some problems we can use symmetry arguments to simplify this dot product and not really do any integration. For an infinite sheet of charge, the surface charge looks the same in all directions, so the field $\mathbf{E}$ must also look the same in all directions. This implies that $\mathbf{E}$ is pointing straight up from the surface and is uniform. In this case, if you use a cylindrical Gaussian surface, the integral simplifies to
$$\oint_{s}\mathbf{E}\cdot d\mathbf{a}=\int E_{z}da=E_{z}\int da=\frac{Q}{\epsilon_{0}}$$
The last surface integral is just the surface area of both end caps $2\pi r^{2}$ and the inclosed charge is simply $Q=\sigma\pi r^{2}$. This leads us to
$$E_{z}=\frac{\sigma}{2\epsilon_{0}}$$

For a finite surface however, this symmetry argument does not hold. The charge distribution does change, specifically at the edge of the surface where it goes to zero. So we cannot be sure of the direction of $\mathbf{E}$ and cannot make any simplifications to the integral.

Why is it that the electric field lines do not remain perpendicular to the surface of the conductor they're coming from if it is finite? Is this just electrostatic repulsion?

 

[NFuller]

Why is it that the electric field lines do not remain perpendicular to the surface of the conductor they're coming from if it is finite? Is this just electrostatic repulsion?

Imagine you pick a point to the right off the edge of the disk. There is now a bunch of charge to the left of the point but none to its right. If the disk is positively charged, this means the electric field will point to the right and not up at this location. The diagram @mishima posted shows this.

 

[Elmer Correa]

Much thanks to you both for the help!