Rank magnitudes of the electron's accerlation.

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Homework Help Overview

The discussion revolves around the acceleration of an electron released between two infinite nonconducting sheets with uniform surface charge densities. Participants are exploring how the electric fields generated by these sheets affect the electron's acceleration and how to rank the magnitudes of that acceleration under different scenarios involving surface charge densities and sheet separations.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relevance of the electric field equations and the effect of sheet separation on the electron's acceleration. There are attempts to calculate the electric fields for different configurations and questions about the directionality of the fields and how they combine.

Discussion Status

Some participants have provided guidance on disregarding the separation of the sheets due to their infinite nature and have clarified how to correctly add the electric fields from both sheets. There is an ongoing exploration of the implications of charge signs and field directions, with some participants expressing confusion about their calculations.

Contextual Notes

Participants are working within the constraints of a homework problem that requires them to rank the magnitudes of acceleration based on given surface charge densities and separations. There is an emphasis on understanding the electric field contributions from both positively and negatively charged sheets.

mr_coffee
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Hello everyone, I'm lost as usual. Because they are two sheets, infinite and nonconducting. I thought I would use this equation: E = \delta/(2Eo). But they give the separation which i don't see how that fits into this equation. I figured I could find the accerlation using F = MA. But anyways, here is the question. An electron is rleleased between two infinite nonconducting sheets that are horiztonal and have uniform surface charge densities \delta(+) and \delta(-). The electron is subjected to the following three situations involving surface charge densities and sheet separations. Rank the magnitudes of the electron's acceleration, greatest first. The answer is they all tie. Here is a link to the image:
http://img333.imageshack.us/img333/4792/image8pt.jpg
Am I approaching this problem totally wrong? :bugeye:
 
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I think you have the right idea. If you know what the electric field of one sheet looks like you'll see that separation is indeed not important. Just add the electric fields from both sheets. Mind the direction.
 
I think you can disregard the separation, as the plates are infinitely long.
 
okay somthing isn't right...
I used E = \delta/2Eo. So I plugged the numbers given and then added up each situation. Like so:

1. E = -4\delta/2Eo; E = +4\delta/2Eo; I added those and got 0.
2. E = -\delta/2Eo; E = +7\delta/2Eo; I added those and got 3\delta/Eo;
3. E = -5\delta/2Eo; E = +3\delta/2Eo; I added those and got -\delta/Eo;

What did i do wrong exactly? :confused:
 
The positive charge does not cancel the negative charge; just the opposite, since they are on opposite sides of the space between the plates. (Remember the parallel plate capacitor? Is the field between the plates zero? No!)

Realize that both the positively charged plate and the negatively charged plate both contribute fields pointing down at any point in between the plates.
 
ohh that's interesting...why would you add them? I see you get the correct answer if you do. I know the E-field goes from + to -. But doens't the negative and + tell the direction or is it just showing the charge?
 
You add the fields because they point in the same direction.

Sure the charge shows you the direction of the field, if you know how to interpret the formula. The field from a plane of charge is E = \sigma / (2 \epsilon_0). When \sigma is positive, the field points away from the plane; when negative, towards the plane.

It does not mean that the field from a positive charge is always "positive", whatever that would mean. If you use signs to indicate the direction of the field, the fields between the plates in this problem are all negative (assuming negative means "down").
 
ohh of course, once i drew the 2 plates and looked at the directions i realized what i did wrong, thanks for the explanation!
 

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