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Electric field of a positive plate

  • Thread starter doggydan42
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  • #1
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1. Homework Statement
There is a large positively charged plate whose charge density is ##\sigma = 2.0 × 10^{-5}\frac{C}{m^2}##. What is the electric field at a point P, that is not enclosed between the plates.

2. Homework Equations
For an infinite sheet:
$$\vec E = \frac{\sigma}{2\varepsilon_0}\hat r$$

3. The Attempt at a Solution
I thought to find the field of the plate by adding, but that would be the field between two infinite planes, one with negative charge and one with a positive charge. Would that still form the electric field of a positively charged plate? So,
$$\vec E = \frac{\sigma}{\varepsilon_0}\hat r$$

Thank you in advance.
 

Answers and Replies

  • #2
TSny
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Hello. Can you clarify the problem statement? The first sentence seems to imply that there is only one plate. The second sentence implies more than one plate.
 
  • #3
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There is one plate. The plate has three dimensions, while the sheet only has two. So there is a thickness to the plate that the sheet does not have.
 
  • #4
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Hello. Can you clarify the problem statement? The first sentence seems to imply that there is only one plate. The second sentence implies more than one plate.
The plate has a thickness, so it is in three dimensions. The sheet is only two dimensional, so it would be a plane in a 3D space.
 
  • #5
TSny
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There is one plate. The plate has three dimensions, while the sheet only has two. So there is a thickness to the plate that the sheet does not have.
So you have a single plate that has some thickness to it? What does it mean when the problem statement says that point P is not between the plates?
 
Last edited:
  • #6
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Well I did not state the full problem, but if it helps here's an image. Part of the problem was finding the electric field to find the force, and once I find that I know how to continue. So the proton is not between the two sheets which form the plates.
 

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  • #7
TSny
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OK, that helps. Point P is not between the faces of the plate. In the problem statement, you say that ##\sigma## is the charge density of the plate. Does that mean that each face of the plate has the surface charge ##\sigma##?

I think it would be helpful to state the relevant part of the problem word for word.
See the paragraph for item 3 here: https://www.physicsforums.com/threads/guidelines-for-students-and-helpers.686781/
 
  • #8
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The whole problem is:
From a distance of 10 cm, a proton is projected with a speed of v = 4.0 × 106 m/s directly at a large, positively charged plate whose charge density is ##\sigma = 2.0 × 10^{−5}C/m^22##. (See below.) (a) Does the proton reach the plate? (b) If not, how far from the plate does it turn around?

So from my understanding, ##\sigma## would be the surface charge of each plate.
 
  • #9
TSny
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Thank you. You have the right idea that you sum the contributions from each surface (sheet of charge). Your expression for E in the "attempt at a solution" is correct, although I'm not sure what the unit vector ##\hat r## denotes?
 
  • #10
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##\hat r## would be the direction of the field.

Thank you.
 
  • #11
TSny
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##\hat r## would be the direction of the field.
And what direction would that be?
 
  • #12
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In this case it would be in the ##-\hat i## direction acting on the proton.
 
  • #13
TSny
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In this case it would be in the ##-\hat i## direction acting on the proton.
OK. Good. I think you have it.
 

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