Electric Field Strength of 10cm x 10cm Charged Copper Plate

[maccha]

A thin, horizontal 10 cm x 10 cm copper plate is charged with 1.0 x 10^10 electrons. If the electrons are uniformly distributed on the surface, what is the strength of the electric field?

I initially went to use the formula given in the text for a conductor where

Electric field=n/Eo

where n is the surface charge density and Eo is the permittivity constant

. I can only get the right answer if I model it as an infinite plane of charge, though. Can you assume that this works even though it's so small?

 

[Redbelly98]

I initially went to use the formula given in the text for a conductor where

Electric field=n/Eo

where n is the surface charge density and Eo is the permittivity constant

That's the equation for the field between a pair of oppositely-charged plates. For a single plate, the field is 1/2 as much.

. I can only get the right answer if I model it as an infinite plane of charge, though. Can you assume that this works even though it's so small?

Yes. You will get the correct E-field right at the surface of the plate. (Moving away from the plate, E will become smaller, unlike the infinite plane case.)

 

[utkarsh1]

Gauss' law works due to symmetry. If the surface is not symmetric, such as 10x10 plate; then you don't know the direction of the electric field so it wouldn't work. (Unless you can use the approximation that distance is very small).
If the surface was infinite plane you know the direction of electric field will be point up/down so you can use gauss law very easily.