Uniform field between parallel plates

In summary: The field from any ring falls off as 1/r2. The field from two parallel plates would then be 1/4th the total field. You would need to do an additional integral to extend the field into a plane.
  • #1
RK1992
89
0
Hey,

I've recently been studying electric fields, and our teacher told us that the field is approximately uniform between two charged parallel plates, ie the force acting on a unit charge is equal everywhere.

This doesn't seem to tie with everything else we've been taught; surely, as you move an electron toward the positive plate, the force of attraction would increase. We've been told to "accept it" without any real explanation, and I'm not seeing the reasoning behind it on my own..

Is anyone able to help me out?

Thanks
 
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  • #2
RK1992 said:
surely, as you move an electron toward the positive plate, the force of attraction would increase.
Why do you think that? Note: Moving towards a plate is not quite the same thing as moving towards a point charge.
 
  • #3
I didn't know that... but why not, surely it's like a line of point charges?
 
  • #4
It is due to the capacitor having a homogeneous electric field. Imagine this:

You have a positive plate - the closer to the plate you place a body with a negative charge, the bigger the attraction force.
Now imagine placing a negative plate opposite to the positive plate that the negatively charged body is between the plates. The negative plate exerts a repulsion force to the body. The closer the body to the negative plate, the bigger the force.

Say the body is exactly in the center of the plates. The attraction and repulsion forces are equal if the plates have equal but opposite charges, and add up.
Now move the body towards the positive plate. The attraction force increases. As it moves away from the negative plate, the repulsion force decreases. The sum of the two stays the same.
 
  • #5
RK1992 said:
I didn't know that... but why not, surely it's like a line of point charges?
No, but you're getting closer! The field from a point charge falls off as one over distance squared, that from a line charge as one over distance, but the field from a charged plane is uniform.
 
  • #6
Doc Al said:
No, but you're getting closer! The field from a point charge falls off as one over distance squared, that from a line charge as one over distance, but the field from a charged plane is uniform.

Is there any way of showing it falls as 1/r ? Seems weird, I don't want to just accept that it does.. I know some calculus, I'm guessing that it will be involved as a sum of infinitely small charges along a line but I'm not sure how to prove it to myself :s
 
  • #7
RK1992 said:
Is there any way of showing it falls as 1/r ? Seems weird, I don't want to just accept that it does.. I know some calculus, I'm guessing that it will be involved as a sum of infinitely small charges along a line but I'm not sure how to prove it to myself :s
You can certainly use calculus to show that the field from a line charge falls off as 1/r (see: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html#c1"). It's a good exercise. (The easy way to show it is to use Gauss's law.)
 
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  • #8
RK1992 said:
But I have no idea why there is a z/r on the end of their equation for the field strength...
They want the component of the field perpendicular to the line (what they call the radial part of the field). The z/r represents cosθ, where θ is the angle that r makes with the normal to the line. (Study the diagram.)
 
  • #9
cdotter said:
This website explains it a little better.

http://faculty.wwu.edu/vawter/PhysicsNet/Topics/ElectricForce/LineChargeDer.html [Broken]

cos(θ) = electric field in the y direction. θ isn't given but cos(θ)=y/r. y is constant (and can therefore be moved outside of the integral). r is found by the pythagorean theorem.

Ah okay, that's a nice explanation :)

Doc Al said:
They want the component of the field perpendicular to the line (what they call the radial part of the field). The z/r represents cosθ, where θ is the angle that r makes with the normal to the line. (Study the diagram.)
Yeah I get it now :D

So I'm now happy with the idea of a line's field varying with 1/r.. how does this tie in with two parallel plates? Is it necessary to do another integral in another direction to extend the line into a plane?
 
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  • #10
RK1992 said:
So I'm now happy with the idea of a line's field varying with 1/r.. how does this tie in with two parallel plates? Is it necessary to do another integral in another direction to extend the line into a plane?
I would consider the field from a ring of charge, then view a plate as a set of concentric rings.
 

1. What is a uniform field between parallel plates?

A uniform field between parallel plates is a region between two flat plates where the strength and direction of the electric field is the same at all points. This means that the electric field lines are parallel and evenly spaced.

2. How is the electric field strength calculated in a uniform field between parallel plates?

The electric field strength in a uniform field between parallel plates can be calculated by dividing the potential difference between the plates by the distance between them. This can also be represented by the equation E = V/d, where E is the electric field strength, V is the potential difference, and d is the distance between the plates.

3. What is the significance of a uniform field between parallel plates in practical applications?

A uniform field between parallel plates is often used in electronic devices such as capacitors and parallel plate capacitors. It is also used in particle accelerators to control the motion of charged particles.

4. How does the distance between the plates affect the strength of the uniform field?

The strength of the uniform field between parallel plates is directly proportional to the distance between the plates. This means that as the distance between the plates increases, the electric field strength decreases.

5. Can a uniform field between parallel plates exist in a vacuum?

Yes, a uniform field between parallel plates can exist in a vacuum. In fact, it is often used in vacuum chambers for experiments and in electronic devices that require a vacuum environment.

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