Looking on Wikipedia, I found the following derivation for capacitance: $$C=\frac{Q}{V}$$ $$V=\int_{0}^{d}E(z)dz$$ It then continues to assume that the electric field lines are straight between the plates (although that's a reasonable assumption), and integrates over the distance between the two plates. It then defines $E$ as being $\frac{\sigma}{\varepsilon}$, all of which I follow. However, the page then goes on to state that the definition of $\sigma$ is $\frac{Q}{A}$, where $A$ is the area of the parallel plate capacitor. My issue is that there are electric field lines from both the top and bottom faces of the parallel plate capacitor. One could argue that there is a greater charge density on the lower side of the plate, and therefore the capacitance contribution from the top side of the plate would be negligible. However, due to the fact that the charge cancels out, the capacitance of the upper face should remain the same regardless of its having a lower charge density.

Another way of illustrating this is to break each plate up into a top face, and a bottom face, and to connect them in parallel - again this would cause a doubling of the area (compared to assuming that both plates simply have only a bottom face, which might be, to say the least, a slight challenge to construct!)

Hence I cannot understand why the capacitance from the top of the plate is not taken into account. Any light you may be able to shed on this issue would be greatly appreciated, if my question is badly worded, I'd be happy to edit elucidate it!

Thanks again!

  • $\begingroup$ I'm not sure I follow, why do you argue that the lower side of the plate has a greater charge density ? $\endgroup$ – HsMjstyMstdn Feb 20 '18 at 17:08
  • $\begingroup$ First you say "the actual charge on the upper face of the capacitor shouldn't make any difference to the capacitance" then you ask why don't we consider the upper face. I'm not sure I follow your reasoning, but it seems you've already answered your own question. $\endgroup$ – The Photon Feb 20 '18 at 17:13
  • $\begingroup$ As for the usual reasoning: The parallel plate formula is derived under the assumption that the plates have infinite extent, so there's no path for field lines from the top surface of one plate to wrap around and terminate on the other plate. $\endgroup$ – The Photon Feb 20 '18 at 17:14
  • $\begingroup$ @HsMjstyMstdn, I wasn't arguing that myself, I was saying that I could see that being a potential argument, and then saying why I thought that it wasn't true, in case therein lay a problem with my reasoning. $\endgroup$ – DoublyNegative Feb 20 '18 at 17:59
  • $\begingroup$ @ThePhoton, I was stating that the fact that there might be a difference in charge between the top and bottom of the plate shouldn't actually affect the capacitance due to the top face of the plate. I'll make that clearer. However, your second point about it being derived from infinitely large plates is pretty much what I was looking for, thanks! (I had a feeling it'd be something like that which I was missing!) $\endgroup$ – DoublyNegative Feb 20 '18 at 18:02

There is no different capacitance from the top of a parallel plate capacitor. The calculation of a parallel plate capacitance, according to your equations, relies on the approximation that the electric field is concentrated only between the plates and that the fringe fields at the edges and the back side can be neglected. Thus the approximated homogeneous field there is simply given by $E=Q/(\epsilon A)$. And the voltage is simply given by $V=E\cdot d$. This approximation works the better the smaller the smaller the plate spacing $d$ , the higher the absolute permittivity $\epsilon$ between the plates, and the larger the area $A$ of the plates is. When you would follow one of the few field lines emanating from the top of a capacitor and ending on the bottom back plate, you would get the same voltage $V$ in the line integral of the electric field. There is no different capacitance for the different capacitor plates with respect to each other. The capacitance of one relative to the other is called the capacitance of the plate capacitor.

  • $\begingroup$ Sorry that I never accepted this answer... Just noticed I hadn't, and your answer explained a lot, so thanks! $\endgroup$ – DoublyNegative Sep 6 '18 at 19:38

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