Gauss Law Anomaly: Potential Across a Capacitor

[rohanprabhu]

I started having this doubt since i saw the derivation of potential across a capacitor using this method. So.. here i ask:

Let there be two large sheets made of insulators, and charged to a planar charge density, \sigma (Sheet 1) and 2\sigma (Sheet 2). Since they are insulators, when placed in proximity to each other, it can be assumed that the charges will not redistribute on either of them.

Now, we put them one against the other with a separation 'd' as is done for a capacitor. Now, i take a cylindrical surface perpendicular to Sheet 1, such that one end of the cylinder is within the thickness of the sheet, with radius 'r'. Then, the flux through the cylinder is only through the other surface. Hence,

<br /> \Phi = E(4\pi r^2)<br />

Using Gauss' law:

<br /> E(4 \pi r^2) = \frac{q}{\epsilon_o}<br />

and hence,

<br /> E = \frac{\sigma}{\epsilon_o}<br />

To calculate the potential, i use dV = E.dl[/tex]<br /> <br /> &lt;br /&gt; V = \frac{\sigma d}{\epsilon_o}&lt;br /&gt;<br /> <br /> However, if i do the same thing but by making the cylindrical surface on the other plate, I&#039;m getting:<br /> <br /> &lt;br /&gt; V = \frac{2\sigma}{\epsilon_o}&lt;br /&gt;<br /> <br /> How am i getting two potential differences for the same system? What am i doing wrong?

 

[Shooting Star]

Now, i take a cylindrical surface perpendicular to Sheet 1, such that one end of the cylinder is within the thickness of the sheet, with radius 'r'. Then, the flux through the cylinder is only through the other surface.

This is not correct. The field inside a conductor is zero, not inside an insulator. So, you cannot say that the flux is only through one surface.

Do one thing -- just take a "plane" charge distribution, so that you don't have to worry about the field inside the "thickness". Now make the same kind of pillbox, and calculate the E, and see the result. For calculating E, you must take both plates into consideration.

 

[rohanprabhu]

This is not correct. The field inside a conductor is zero, not inside an insulator. So, you cannot say that the flux is only through one surface.

thanks for clearing that up!

For calculating E, you must take both plates into consideration.

But if i take a pillbox, like that, then how can the other plate have any effect since the charge resides outside the Gaussian surface?

Plus, the direction of electric field will be opposite on both the faces of the cylinder? Shouldn't this make the flux zero? I think I'm terribly confused on this one.

 

[harshant]

To start with, let me consider only one "plane " charge distribution (without thickness as Shooting Star mentioned). We are certain that the field is same on both sides of this plane, because of symmetry. Now we can simply apply Gauss law and get the field as has been mentioned.

But when we have two planes instead of one, although the flux through the pillbox remains unchanged (because the charge inside the pillbox is same), we cannot apply the same symmetry argument as before concerning the fields on either side of one of the planes. This is because in this situation we have another plane on one side but the other side remains unchanged. Therefore the field is not the identical on both sides of the plane as before. This is how the other plane exerts its "effect". I hope this clears your first doubt.

As for your second doubt, you are right that direction of electric field will be opposite on either side but so will be the direction of the "outward" normals that we consider for calculating the flux. Therefore the flux will be outward in both directions (scalar product is positive on both faces) and the fluxes will add rather than cancel out.

 

[Shooting Star]

But if i take a pillbox, like that, then how can the other plate have any effect since the charge resides outside the Gaussian surface?

Plus, the direction of electric field will be opposite on both the faces of the cylinder? Shouldn't this make the flux zero? I think I'm terribly confused on this one.

That was a slip on my part, asking you to take both plates into consideration to calculate E. It complicates matters. So, just to be on the safe side, let's review the method of finding the field of one infinite charged plane, which you probably know very well. More or less the same thing has been written by harshant.

Consider just the standard infinite plane charge distribution \sigma.

Take a cylindrical pillbox, whose flat faces are at equal distance from the plane. The field has to be normal to the plane, by symmetry, and in opposite directions on opposite sides, and of equal magnitude at the same distance from the plane. The flux is the integral E.da over the flat surfaces. E is perpendicular to n on the curved surface, and parallel to the unit normal n on both flat surfaces. So, on both the flat surfaces, the flux E.da is nothing but E*pi*r^2, and so we can find E at any distance from the plane. It turns out E is independent of the distance from the plane. So, the field is constant, but with different signs on opposite sides of the plane. All this, I'm sure you know -- it's standard textbook stuff.

If another charged plane parallel to the first one is there, we'll simply apply the principle of superposition to find the E at any point. If \sigma and -\sigma are the surface charge densities, as in capacitors, then in the middle of the two plates, the field is twice that of due to one plate with \sigma. On the outer sides, the E of one plate cancels the -E due to the second plate.

This is exactly what happens in a parallel plate capacitor, where you have a non-zero field only between the plates. So, V/d=E, since E is constant between the plates.

In the example you have given, find the fields due to the individual plates, and then add them, as vectors, to get the resultant field at any point.