Principle of Superposition for Electric field

[hangainlover]

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. .

 

[Zacku]

Hi,

I am not sure I understand all your questions but about the capacitor you can't simply apply the superposition property from the result of two non-metallic isolated plates.

If you have a plate capacitor with metallic plates, the resulting field inside the slit can be in principle understood as the superposition of each field from the charged layers (non-metallic if you want) plus the corresponding responses from the metallic material they are made of.

This response from the medium is a priori non trivial but the whole system will attain a steady situation eventually where it can be shown that this complex answer can be accounted for by requiring the metallic boundary condition you mentioned i.e. normal outward electric field equals \sigma/ \epsilon_0.

That's what you do when solving the electrostatic problem for a plate capacitor.

 

[willum97]

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?

No you don’t. (sorry for the late reply, hopefully your still about)

If you have a single conducting plate and give it a charge Q/2 then this charge will be distributed over the 2 surfaces. In absence of any other charges, the field lines will be extended into space. However you could just as well draw a conducting sphere having a huge radius and ending the lines on the inside of this sphere. These lines will be ending on charges of opposite sign then the ones on the conducting plate.

So what I am saying is, is that I have now introduced an equal and opposite charge without adding anything to the original charge. Wow a single charge is never really a single charge it always implies a double.