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I am doing an investigation into the differences of calculating capacitance using the well know formula for an idealistic parallel plate capacitor, based on the assumption of a uniformly distributed electric field:$C=\epsilon_0 \frac{A}{d}$ vs using numerical methods for calculating the capacitance of a realistic model i.e. with fringe fields (see here for a more detailed explanation of my methods).

I am investigating how the capacitance changes as i move the plates further away from each other, and what i have found is that, based on the results I'm getting, as the plates are moved further away from each other the ratio of realistic/idealistic increases, implying that the realistic electric field can store more energy.

My question is, is this correct and if not why?

Thanks

[Edit] I originally thought that this may be because the realistic formula is only valid for small d, however the ratio seems to increase linearly, rather than converging to a value and then dropping off leading me to think that this is not the problem?

[Info for comments] $$C=\frac{\epsilon_0 L}{V} \sum_{bound} |\phi_{outer}-\phi_{plate}|$$

$$C_{\infty}=\frac{\epsilon_0 A}{d}$$

Dividing them, where $A=lL$ i.e the area of the plate, gives:

$$\frac{C}{C_{\infty}}=\frac{d}{V l} \sum_{bound} |\phi_{outer}-\phi_{plate}|$$

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  • $\begingroup$ Why do you say that if the ratio of realistic/idealistic increases, that implies that the realistic electric field can store more energy? $\endgroup$
    – cinico
    Commented Dec 8, 2013 at 22:32
  • $\begingroup$ When the plates are 'pulled' apart the ratio goes up... surely this means that the numerator is either increasing faster or decreasing slower than the denominator implying that the capacitance based on the realistic field is higher? $\endgroup$
    – WireInTheGhost
    Commented Dec 8, 2013 at 22:44
  • $\begingroup$ Ok, so according to your edit this ratio does not converge to 1, right? $\endgroup$
    – cinico
    Commented Dec 9, 2013 at 10:42
  • $\begingroup$ Can you put in more detail how did you derived the ratio formula? I don't see how does the length of the plates appear. $\endgroup$
    – cinico
    Commented Dec 9, 2013 at 10:45
  • $\begingroup$ No it does not converge at all. The derivation is quite long, is there anyway i can upload a file instead of making my question a few pages long? Essentially it used the taylor expansion of a derivative to express E=nabla(phi) as a difference of potential. This is substituted into a formula derived from Maxwells equation, where the integral is replaced with a sum, as this os over a finite number of points. This give the formula for C i have included above. Once divided by the idealistic capacitance the l drops out as shown above. $\endgroup$
    – WireInTheGhost
    Commented Dec 9, 2013 at 12:00

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The solution for the plate capacitor is an approximation for separations much smaller than the plate area scale. Therefore it will be unusable if the distance between the plates becomes large.

One thing the approximation does not take into account is that even if the distance between the plates goes to infinity or one plate is removed, one plate will still have a finite capacitance against the zero potential in infinity.

It is therefore to be expected, that the error is of order $$ \frac{C}{C_\infty} \sim d $$

You should be able to easily check wether I am right. You could include some example data in your question ($d$ against $C$ and $C_\infty$).

However I don't fully get how you arrive at your numerical result for $C_\infty$.

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