Charged Conducting Sheet v. Charged Non-Conducting Sheet

[Taulant Sholla]

Homework Statement

An electron is shot directly toward the center of a large metal plate that has surface charge density sigma. If the initial kinetic energy of the electron is known and if the electron is to stop (due to electrostatic repulsion from the plate) just as it reaches the plate, how far from the plate must the launch point be?

Relevant Equations

E=sigma/a/epsilon_0

The solution to this problem states the electric field is E=σ/ε₀. Is that because it's a conducting plate? I know for a non-conducting plate it's E=σ/2ε₀. This is a Gauss' Law problem. I know how to derive for non-conducting plate. What's different with conducting plate derivation? Thank you!

 

[haruspex]

What's different with conducting plate derivation?

For a non-conducting plate, there need not be any other charge on it, but for a conducting plate there is necessarily an equal charge on the further surface.

One thing bothers me about this question, though: as the electron approaches the plate, the charges on the plate will redistribute.

 

[Delta2]

One thing bothers me about this question, though: as the electron approaches the plate, the charges on the plate will redistribute.

Strictly speaking you are right but for the level of this problem (i suspect high school or college level) we can safely consider as negligible the EM field produced by the moving electron (or just neglect the E-field in the quasi static approximation).
Does it have analytical solution if we don't neglect the EM field produced by the electron? I don't think so.

 

[Isaac0427]

For a conducting plate in electrostatic equilibrium such as the metal one in your problem, the electric field indeed is $\sigma /\epsilon_0$. To derive this, one crucial thing has to be considered-- the electric field inside of a conductor in electrostatic equilibrium is zero (hopefully you know the qualitative argument as to why this is!) Set up a Gaussian cylinder with surface area of the faces each being A, and have one end of the cylinder be inside the conductor and the other outside (this is the same Gaussian surface used in the non-conducting plate derivation). Using the crucial fact mentioned above, do you see where you lose that factor of 2 in the denominator of $\sigma /2\epsilon_0$?

 

[haruspex]

Strictly speaking you are right but for the level of this problem (i suspect high school or college level) we can safely consider as negligible the EM field produced by the moving electron (or just neglect the E-field in the quasi static approximation).
Does it have analytical solution if we don't neglect the EM field produced by the electron? I don't think so.

Yes, I'm sure it's fine as long as the initial distance from the plate is large compared with the average spacing of electrons on the plate.

 

[Delta2]

Yes, I'm sure it's fine as long as the initial distance from the plate is large compared with the average spacing of electrons on the plate.

i am not so sure i understand you here, can you expand?

 

[haruspex]

i am not so sure i understand you here, can you expand?

Under that condition, for most of the journey of the electron towards the plate there will be very little redistribution.

 

[Delta2]

Under that condition, for most of the journey of the electron towards the plate there will be very little redistribution.

Yes, but since the stopping point of the electron is just above the plate, there will be some fraction of the journey (when the electron gets very close to the plate) where the redistribution is not negligible, but yes this is a tiny fraction of the total journey of electron.

 

[haruspex]

Yes, but since the stopping point of the electron is just above the plate, there will be some fraction of the journey (when the electron gets very close to the plate) where the redistribution is not negligible, but yes this is a tiny fraction of the total journey of electron.

Right, but what matters is how that compares with the initial spacing of the electrons. E.g. start with just five electrons in a square sheet. They will be one at each corner and one central. When the fired electron reaches the sheet, the fired electron will have deviated one way and the central electron another.

 

[Delta2]

Right, but what matters is how that compares with the initial spacing of the electrons. E.g. start with just five electrons in a square sheet. They will be one at each corner and one central. When the fired electron reaches the sheet, the fired electron will have deviated one way and the central electron another.

Yes , the way i understand it is that when the distance of the electron is big, in comparison with the inter-electron spacing in the plate, then the force from the electron will be small in comparison with the interelectron forces in the plate, hence it won't affect much the positioning of electrons in the plate.