Principle of Superposition for Electric field

In summary, the charge on a single conducting plate is evenly distributed over the two surfaces. The electric field outside the plate is the superposition of the field from the charged layers and the response from the medium. This response from the medium is non trivial but the whole system will eventually attain a steady situation where it can be shown that this complex answer can be accounted for by requiring the metallic boundary condition.
  • #1
hangainlover
83
0
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. .
 
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  • #2
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 [tex]\sigma/ \epsilon_0[/tex].

That's what you do when solving the electrostatic problem for a plate capacitor.
 
  • #3
hangainlover said:
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.
 

1. What is the Principle of Superposition for Electric field?

The Principle of Superposition for Electric field states that the total electric field at a point in space is equal to the vector sum of the electric fields produced by individual charges at that point.

2. How does the Principle of Superposition apply to multiple charges?

For multiple charges, the Principle of Superposition states that the total electric field at a point is equal to the sum of the electric fields produced by each individual charge. This holds true for any number of charges, as long as they are all in the same region of space.

3. Can the Principle of Superposition be applied to both positive and negative charges?

Yes, the Principle of Superposition applies to both positive and negative charges. The electric field produced by a positive charge is directed away from the charge, while the electric field produced by a negative charge is directed towards the charge. When vectorially adding these fields, the direction of the resulting electric field will depend on the relative positions and strengths of the charges.

4. How does distance between charges affect the electric field according to the Principle of Superposition?

The Principle of Superposition states that the electric field at a point is directly proportional to the magnitude of the charge and inversely proportional to the square of the distance between the charge and the point. This means that as the distance between charges increases, the electric field strength decreases.

5. Can the Principle of Superposition be applied to continuous charge distributions?

Yes, the Principle of Superposition can be applied to continuous charge distributions by breaking the distribution into small enough pieces that they can be approximated as point charges. The electric field at a point due to a continuous charge distribution is then calculated by summing the individual electric fields produced by each small piece of the distribution.

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