Capacitance from both sides of a parallel plate capacitor

Question

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!

Answer 1

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.