# Perpendicular charged plates

1. Feb 25, 2004

### SparkimusPrime

Excerpt from:

Beicher and Serway, "Physics for Scientists and Engineers with
Modern Physics, 5th edition"

Chapter 24, problem 58, part c:

Two infinite, non conducting sheets of charge are parallel to each other, as show in the figure. The sheet on the left has a uniform surface charge density A, and the one on the right has a uniform charge density -A. Calculate the value of the electric field at points...

(c) to the right of the two sheets.

The solution is supposed to be zero, but I'm not sure why. It seems that there should be some electric field to the right of the plates due only to the negatively charged plate. Taking the Gaussian approach and enclosing them in a sphere seems to suggest that there is no net electric charge outside the two plates, because the charges are equal and opposite, but I'm not sure. Would someone care to give me another point of view?

#### Attached Files:

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2. Feb 25, 2004

### cookiemonster

How exactly are you going to manage to fit an infinite plane sheet into a sphere? Why don't you try putting it in a nice little pillbox instead, and then let the dimensions of the pillbox go to infinity?

Here's an intuitive little hint, though: Does the view of the infinite sheet ever change depending on how far away from it you get?

cookiemonster

3. Feb 25, 2004

### paul11273

Sparkimus,
In your drawing, it looks like you are only showing the field due to the negative plate. What about the positive plate?

I think that when you draw the field of the positive plate, you will see that in between the plates the field is heading in the same direction, and so they add. On either sides, the fields are opposite, and so will cancel.

Part a) of the question asked for the field to the left of the positive plate. Did you get that one correct, and understand why? If so, the answer is the same for the field to the right of the negative plate, and for the same reason.

4. Feb 25, 2004

### gnome

Elaborating a little on cookiemonster's remark:

It's not so much a question of how can you fit an infinite sheet into a sphere. The crucial requirement is symmetry. For example, if you have a finite charged disk you can always imagine a sphere big enough to completely enclose it but that sphere would NOT be a Gaussian surface because all points on the disk would NOT be equidistant from every point on the surface of the disk, and therefore the electric field would vary over the surface.

Re-read items 1-4 near the top of page 750 of Serway.

Also, review Example 24.8 in the text, where they explain the use of a "Gaussian cylinder" to analyze continuous sheet charges, where they show why the field outside an infinite charged plane is given by
$$E = \frac{\sigma}{2\epsilon_0}$$
Notice in particular that there is no "x" (or "r") in that expression. That is, from a single charged plane the electric field is the same regardless of distance from the plane.

Now, if you understand the reasoning behind the Gaussian cylinder for one charged plane, imagine the same Gaussian surface penetrating through your two parallel planes. The simple Gaussian argument is that since one plane is + and the other is -, the net charge enclosed by the cylinder is 0.
Therefore the net flux through each end is 0 and therefore the field at each end is 0.

Another way to look at it, if you remember that the magnitude of these electric fields is constant regardless of distance from the plane: in problem 58 the charge distributions are equal in magnitude but opposite in sign. So at any point in space to the right of BOTH planes, there is an electric field &sigma;/(2&epsilon;0) from one plane, and a field -&sigma;/(2&epsilon;0) from the other. So you have two vectors of equal magnitude but pointing in opposite directions and their sum is 0. The same applies on the left of both planes.

But in between the two planes, a vector pointing OUT from the positive plane and a vector pointing IN to the negative plane are both pointing in the same direction. This is why, as paul11273 pointed out, the fields add only in that region.

Last edited: Feb 25, 2004
5. Feb 25, 2004

### SparkimusPrime

paul11273:

I was thinking in straight lines, excuse me. From my understanding the negative plate would not put forth an electric field, it only receives electric field "lines." But from what you said I think I have an inkling of what you mean. You're saying that the electric field from both plates expands outward through spaces like ripples in a pond. One splash in the (hypothetical) pond would produce concentric waves expanding outward, losing intensity the farther they move from the center of the splash. While two such splashes would add disturbance in the middle and interfere with each other around the edges. Is that correct?

I wish I had a scanner or a tablet, expressing my ideas in solely in words is frustrating.

No I didn't get this one right, I misinterpreted the question, but with this insight I'll redo it.

Gnome:

Textbooks don't do a thing for me, never have. I absorb knowledge mainly by solving a problem with someone, or reading someone's personal input about a concept. Which is why, I suppose, this class is kicking my butt so hard, my professor is brilliant; brilliant to the point that she covers concepts so rapidly I find it difficult to digest the full import of her meaning. Very frustrating, this helps a great deal. Thank you.

Back to the problem at hand. I was under the, apparently, mistaken impression that electric fields end at an equal and opposite charge as the originating object. This of course, in retrospect, is silly. I'll rethink the problem with these insights. Thanks again.

Last edited: Feb 25, 2004
6. Feb 25, 2004

### gnome

This "ripple" idea bothers me a little. Don't lose sight of the fact that, if the charged surface is a flat plane extending "infinitely" in all directions, the NET electric field at every point outside the plane is a vector perpendicular to the plane. If you're thinking of the field as emanating from individual charged points on the plane, any component of field emitted at an angle to the plane other than perpendicular is opposed by one emitted from another point at an "opposite" angle, so the net effect is ONLY perpendicular.

7. Feb 25, 2004

### paul11273

Spark,
You are correct about the negative plate accepting field lines.
And the positive plate sends out field lines.

See my attachment. I have the next edition (6th?) of the same book, so my note on which problem number and pg for equation etc is a little different.

Take a look at the diagram I drew, with field lines from the positive and to the negative plates. This should help you to "see" the cancellation on each side of the plates. Notice in the middle the fields are heading in the same direction, that is why they add.

Let me know if this helps.

8. Feb 25, 2004

### paul11273

Here is the attachment...I hope it is readable. I had to scan it about 4 times at lower resolutions and then make it smaller the 400x400.

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9. Feb 25, 2004

### paul11273

Spark,
I just looked at the attachment and it came out pretty crappy.
If you want, PM me with you email address, and I will send you a good version of the scan.

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