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If a parallel plate capacitor is formed by placing two infinite grounded conducting sheets, one at potential $V_1$ and another at $V_2$, a distance $d$ away from each other, then the charge on either plate will lie entirely on its inner surface. I'm having a little trouble showing why this is true.

In the space between the two plates the field $E = ( V_1 - V_2 ) / d$ satisfies Laplace's equation and the boundary conditions, from which I can derive the surface charge density is $\pm E / 4 \pi$. But how about the space above and below the capacitor? Certainly I can't just use superposition of the inner surface charge distributions to say that the field outside the capacitor is zero, (and thus the surface area charge density is zero), for this assumes there is no charge on the outer surfaces to begin with.

Any help clearing up this mental block would be greatly appreciated, thanks.

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Off the bat, I would treat this problem as unsolvable because there's no such thing as an infinite capacitor, and even if one did exist, it could never be charged. Now, saying that a capacitor's radius (assume a circular plate...if it's big enough its shape doesn't really matter) compared to the plate separation is large is a different, yet much more realistic, way of characterizing the capacitor. –  user11266 Jan 22 '13 at 0:50
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3 Answers

One could deal with the problem by being careful with how one constructs a mathematical interpretation of the physical system. I will treat the simplest case: treat the surfaces of the parallel plate capacitors as true two dimensional surfaces. In this case there is no inner or outer surface charge, just a surface charge density defined on each surface.

Mathematically one could represent each conductor as an infinite plane, say $S_\pm \subset \mathbb{R}^3$, then there are two surface charge densities $\sigma_\pm$ each defined on the corresponding surface $S_\pm$. Alternatively, one may use the language of distributions and use a (volume) charge distribution defined on all of $\mathbb{R}^3$ such that $\rho(x, y, z) = \sigma_+ \delta(z - d/2) + \sigma_+ \delta(z + d/2)$ where I have put $S_\pm$ on the planes $z = \pm d/2$.

More complicated models might assume each plate of the conductor has a finite thickness. One could then solve the more complicated problem and compute what happens in the limits at the thickness approaches zero.

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First, a nit: if the potentials on the two plates are non-zero they are not grounded, by definition.

Second, the way I think about it: in the region of interest above and below the plates, the boundary conditions are not established. To set these boundary conditions, you could imagine adding an additional two infinite conducting plates above and below the original plates, and grounding these new plates to 0 potential.

  • If the new plates are initially located close to the original plates, there will indeed be an electric field above and below the original plates, and a corresponding surface charge density on their outside surfaces.
  • Now imagine the new plates being removed to infinity. Since the potential differences are fixed, the electric field, and the outer surface charge densities, go to zero.
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So the reason that setting the potential at infinity to zero would lead to zero surface charge is because the field for an infinite sheet is essentially constant? Also, if the plates were truly infinite, how would we deal with the problem? At that point talking about zero potential doesn't mean anything. –  Blooper Mar 27 '12 at 17:59
    
Thinking about the problem, it also seems like having the same charge density on the outer surface of each plate should be a solution, because then the field inside the conductor would still be zero. –  Blooper Mar 27 '12 at 18:05
    
To the first comment, yes, with these idealized infinitely long and wide sheets, the E-fields are just the potential difference / sheet separation (solution of Laplace's equation in a very simple geometry). I don't see any problem with grounding arbitrarily large plates. –  Art Brown Mar 27 '12 at 18:17
    
To the second comment, such an outside surface charge density would create an electric field above and below the plates, running off to infinity, that would raise the plate potential to infinity, violating the problem conditions. (In other words, this field isn't consistent with the boundary conditions.) –  Art Brown Mar 27 '12 at 19:03
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Ignore inner and outer surfaces. There is just one surface.

Imagine a single, infinite plane with some positive charge density. You can easily show there would be an electric field of constant strength*, perpendicularly out of the plane all the way to infinity on both directions.

Now imagine a single, infinite plate with the same negative charge density. There would be an electric field of constant strength perpendicularly into the plane all the way to infinity in both directions.

Put these two plates on top of each other, and these fields perfectly cancel.

Put these two plates in parallel, and because the field is constant strength it will perfectly cancel everywhere except between the two plates, where the electric field directions are the same and it will add to be twice as strong.

[*By constant strength I mean the electric field is just as strong no matter how far you are from the plate. Why is the field constant strength? Because the field lines can't ever diverge from one another. The way fields usually get weaker is the equipotential surface the field lines are normal to gets bigger as you increase the distance from the object. So the same number of field lines piercing a bigger surface means a field lines are more spread out, and thus a weaker field. In this case however, the equipotential surfaces are always a pair of infinite parallel planes, no matter what distance we are from the charged plane. No spreading means no change in field strength.]

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