What is the Electric Field Between Two Charged Plates?

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SUMMARY

The electric field between two parallel charged plates can be calculated using the formula E = sigma/epsilon, where sigma is the surface charge density and epsilon is the permittivity of free space. In this discussion, the plates have a charge of +2.70 x 10^-3 C and are separated by a distance of 2.20 cm, with a side length of 0.820 m. The separation distance d is negligible when compared to the plate dimensions, thus it does not affect the electric field calculation. The correct value for epsilon is 8.85 x 10^-12 F/m, which is crucial for accurate results.

PREREQUISITES
  • Understanding of electric fields and charge distributions
  • Familiarity with the concept of surface charge density (sigma)
  • Knowledge of the permittivity of free space (epsilon)
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the derivation of the electric field between parallel plates
  • Learn about the implications of plate separation on electric field strength
  • Explore the concept of capacitance in parallel plate configurations
  • Investigate the effects of different dielectric materials between the plates
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Students and professionals in physics, electrical engineering, and anyone interested in understanding electrostatics and electric fields in parallel plate systems.

lobstrain
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Alright, here's the problem. You're told that there are two square metal plates with side length L and a distance d away from each other. One has charge +Q, the other -Q. Then they ask for the magnitude for the charge between the plates, not close to the edge.

Here's what I have so far: The electric field between them obviously isn't 0.

I've got sigma = (magnitude of charge)/area.
The electric field for one plate is E = sigma/(2 * epsilon).
Since the fields from both plates in between them point in the same direction, the total field would be E = sigma/epsilon.

This was going great until I realized the I still had an unused variable, distance d. I have no idea of what to do with it. I tried multiplying E by d, then dividing it by d, but I don't like the way those numbers look. Please help!
 
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Separation distance d is immaterial to the electric field if d is small compared to L. Does the question go on to ask about the capacitance?

AM
 
Nope. Correct me if I'm wrong, but this is what I have pictured (since there is no picture to go with the problem):

| - ---> +
| - ---> +
| - ---> +
L - <-d-> +
| - ---> +
| - ---> +
| - ---> +

For the record, L = 0.82m and d = 0.022m

I'm not sure if that would be considered a significant difference, but they gave me a value, so I'm scared to ignore it.
 
Last edited:
bu-du-bump!
 
I'm sorry,but to me the problem does not make too much sense.What's inbetween the plates??Vacuum,a conductive medium,a dielectric,what??
Besides,how would u get the charge?

Daniel.
 
Ok, I'll put it verbatim.

Two square metal plates are placed parallel to each other, separated by a distance d = 2.20 cm. The plates have sides of length L = 0.820 m. One of the plates has charge Q = + 2.70 x 10^-3 C, while the other plate has charge -Q. What is the magnitude of the electric field between the plates, not close to the edge?
 
lalalabump
 
Better listen to Andrew and ignore that separation distance. Note that the equation you are using for the field:
"The electric field for one plate is E = sigma/(2 * epsilon)."​
is only accurate for distances that are small compared to L. Use it. That's what they want.
 
lobstrain said:
Ok, I'll put it verbatim.

Two square metal plates are placed parallel to each other, separated by a distance d = 2.20 cm. The plates have sides of length L = 0.820 m. One of the plates has charge Q = + 2.70 x 10^-3 C, while the other plate has charge -Q. What is the magnitude of the electric field between the plates, not close to the edge?

That's something totally different.Here's something from your first post: "Then they ask for the magnitude for the charge between the plates, not close to the edge. "... :rolleyes:

Daniel.
 
  • #10
Hahaha, I was doing it right the whole time. The only thing was that I kept putting 8.55 E -12 for epsilon instead of 8.85. Thanks guys!
 

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