High School Curious about the uniform electric field between metal plates

[hongseok]

TL;DR

How is there a uniform electric field between two parallel metal plates? Okay, I can understand it through electric field lines. But how can this be quantitatively confirmed? My intuition is that as the charge approaches the metal plate, r becomes very small, so according to Coulomb's law, a much larger electric field must be applied. Is it perhaps canceled out by the assumption that it is an infinite sheet of metal? Please provide detailed explanation.

How is there a uniform electric field between two parallel metal plates? Okay, I can understand it through electric field lines. But how can this be quantitatively confirmed? My intuition is that as the charge approaches the metal plate, r becomes very small, so according to Coulomb's law, a much larger electric field must be applied. Is it perhaps canceled out by the assumption that it is an infinite sheet of metal? Please provide detailed explanation.

 

[renormalize]

TL;DR Summary: How is there a uniform electric field between two parallel metal plates? Okay, I can understand it through electric field lines. But how can this be quantitatively confirmed? My intuition is that as the charge approaches the metal plate, r becomes very small, so according to Coulomb's law, a much larger magnetic field must be applied. Is it perhaps canceled out by the assumption that it is an infinite sheet of metal? Please provide detailed explanation.

Are you asking about the uniform field inside a parallel-plate capacitor in the electrostatic case? If so, there is no magnetic field involved whatsoever.

 

[kuruman]

My intuition is that as the charge approaches the metal plate, r becomes very small, so according to Coulomb's law, a much larger magnetic field must be applied. Is it perhaps canceled out by the assumption that it is an infinite sheet of metal?

I assume you wrote "magnetic" when you actually meant to say "electric" field.
What charge approaches the metal plate? Are you charging a parallel plate capacitor? The electric field is approximately uniform between the plates. Most notably, there is fringing at the edges.

 

[Baluncore]

The electric field between parallel plates is assumed uniform because the constant separation between the plates is small, when compared to the extent of the plates.

While changing voltage and charge, the plates of a parallel capacitor, can be modelled as a transmission line. During a change of voltage, there will be a temporary magnetic field associated with the charge that propagates.

 

[hongseok]

I assume you wrote "magnetic" when you actually meant to say "electric" field.
What charge approaches the metal plate? Are you charging a parallel plate capacitor? The electric field is approximately uniform between the plates. Most notably, there is fringing at the edges.

Yes, I wanted to talk about electric fields. Sorry

 

[kuruman]

Yes, I wanted to talk about electric fields. Sorry

Has your question been answered?

 

[sophiecentaur]

My intuition is that as the charge approaches the metal plate, r becomes very small, so according to Coulomb's law, a much larger electric field must be applied.

I think your intuitive mental model is based on an array of charges spaced all over the plate. Each charge will contribute to the force on your test charge. As the test charge is brought closer: 1. The nearby charges will cause a greater and greater charge in the Normal direction. 2. Charges elsewhere on the plate (away from the normal) will produce Coulomb's Law forces but symmetry will cause the tangential component of forces to cancel. The Normal componens of those forces will tend to zero. Only the charge directly under the test charge will produce a net Normal force. So the net Coulomb's Law forces will remain the same for all distances from the plate.
That's what the Maths says, too.

 

[kuruman]

Test charges by definition "test" an existing electric field without affecting the charge distribution that produces this field.
https://en.wikipedia.org/wiki/Test_particle

 

[A.T.]

My intuition is that as the charge approaches the metal plate, r becomes very small, ...

What also becomes very small is the amount of charge (or area) on the plate, within any given angle range, off the normal vector, which determines the relevant normal component of the Coulomb forces by those charges.

Since that area is proportional to r², while the Coulomb force is proportional to 1/r², the two effects cancel.