- #1
hangainlover
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This has been bothering me for a long time and I can't find an adequate explanation..
Lets say you have an infinite conducting plate with finite thickness D and it is given charge Q
the charge will be equally split into two and concentrated only on the two surfaces.
Each surface will get Q/2
and the magnitude of electric field outside the plate is E=Q/(2A*epsilon naught) (E for a conducting plate is surface charge density/epsilon naught as opposed to surface charge density/(2epsilon naught)
Then, if you have two nonconducting plates separated by distance D and each plate gets Q/2 which is uniformly distributed ..
then outside the space between the two plates, E = 2*(Q/4A*epsilon naught)
But why can't we use superposition principle in the case of one conducting plate with Q and say you get E=Q/(A epsilon naught)i.e. Q/(2A epsilon naught) from each surface...
Is it because we are looking at the mid section of the conductor as some kind of Faraday cage? (since the plate is essentially infinity and blocks the electric field from the other side?)
Also, let's look at two parallel plate capacitor..
One plate gets +Q/2 and the other gets -Q/2
I know the formula says E= charge density/epsilon naught.. so E has to be Q/(2A epsilon naught) However, it seems rather inconsistent in the case of the parallel plate capacitor...
We know all the charge is going to be concentrated on the inner surface
So shouldn't you get E = Q/(2A epsilon naught) solely from one plate and since you have TWO plates!, shoudlnt you multiply Q/(2A epsilon naught) by 2?
I mean.. I can derive E field inside a parallel plate capacitor from Q=CV and confirm that E field = Q/(A*epsilon naught) ...
But it seems to me that you get the same value from the other plate..so shouldn't you double it?
Thanks. .
Lets say you have an infinite conducting plate with finite thickness D and it is given charge Q
the charge will be equally split into two and concentrated only on the two surfaces.
Each surface will get Q/2
and the magnitude of electric field outside the plate is E=Q/(2A*epsilon naught) (E for a conducting plate is surface charge density/epsilon naught as opposed to surface charge density/(2epsilon naught)
Then, if you have two nonconducting plates separated by distance D and each plate gets Q/2 which is uniformly distributed ..
then outside the space between the two plates, E = 2*(Q/4A*epsilon naught)
But why can't we use superposition principle in the case of one conducting plate with Q and say you get E=Q/(A epsilon naught)i.e. Q/(2A epsilon naught) from each surface...
Is it because we are looking at the mid section of the conductor as some kind of Faraday cage? (since the plate is essentially infinity and blocks the electric field from the other side?)
Also, let's look at two parallel plate capacitor..
One plate gets +Q/2 and the other gets -Q/2
I know the formula says E= charge density/epsilon naught.. so E has to be Q/(2A epsilon naught) However, it seems rather inconsistent in the case of the parallel plate capacitor...
We know all the charge is going to be concentrated on the inner surface
So shouldn't you get E = Q/(2A epsilon naught) solely from one plate and since you have TWO plates!, shoudlnt you multiply Q/(2A epsilon naught) by 2?
I mean.. I can derive E field inside a parallel plate capacitor from Q=CV and confirm that E field = Q/(A*epsilon naught) ...
But it seems to me that you get the same value from the other plate..so shouldn't you double it?
Thanks. .
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