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Suppose there is a parallel-plate capacitor but the two plates have different areas, $A_1$ and $A_2$. How will we derive an expression for its capacitance?

I have been told to take the area common between the two, but from where does this follow?

Ideally we start off by assuming that one plate has $+Q$ charge and the other has $-Q$. Then we find that the potential difference $V$ is directly proportional to $Q$, from which we find capacitace $C$.

I am unable to find the potential difference.

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The capacitance is a result of the polarization of the medium due to electric field and the attraction of charges on one plate due to the charge on the other (as mediated by the electric field).

When you have two plates facing each other, the electric field is present in their common area (ignoring small fringe effects).

This is why you use the area of overlap to compute the capacitance.

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    $\begingroup$ I think this is correct explanation as electric field of plane sheet is perpendicular to surface right? Thank you a lot. $\endgroup$ – Max Payne Jan 4 '17 at 15:00
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    $\begingroup$ What if there is vacuum between the two plates? I would think that using the overlap area would be a poor approximation in that case, wouldn't it? $\endgroup$ – garyp Jan 4 '17 at 15:05
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    $\begingroup$ @garyp - no, the force of attraction of the charges of one plate on charges in the other plate rapidly fall off when you move away from the area of overlap. The approximation will only break down if the ratio of spacing to lateral dimension is not small (that is, when the gap is "large "compared to the size of the plate) - in that case edge effects are not insignificant (but also, the usual parallel plate equation is no longer accurate). $\endgroup$ – Floris Jan 4 '17 at 15:07
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    $\begingroup$ Thanks. Actually, the picture I had in my head was the large-gap case. I was thinking of the general problem, not the practical problem. :) $\endgroup$ – garyp Jan 4 '17 at 15:10
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In general, the areas of the plates add like $(\frac{1}{A_1}+\frac{1}{A_2})$, as I will prove. The capacitance reducing to "taking common area" only in the limit of one of the plates being much larger than the other.

$$C=\frac{Q}{|\Delta V|}$$

WLOG, let $A_1\geq A_2$, let the plates be separated by a distance $d$, and suppose that $|Q|$ is on both plates. Now,

$$\Delta V=-\int_0^d \mathbf{E}\cdot d\mathbf{l}.$$

Where $\mathbf{E}$ is defined to be the electric field between the plates. Now, the electric field between two finite plates can get quite complicated if we are not armed with the assumption $$d<<A_1,A_2.$$

Given that, the electric field between the plates can be taken to be uniform. Now, use Gauss' law (using the "pillbox" surface). Since $A_1$ is bigger by assumption, the electric field between the plates, then, is determined by

$$E_1\cdot A_{pillbox}=E_1\cdot 2 A_{circle} =\frac{1}{\epsilon_0}\sigma_1 A_{circle} $$

Therefore,

$$E_1=\frac{\sigma_1}{2\epsilon_0}=\frac{Q}{2A_1\epsilon_0}. $$

For, the other plate we have that

$$E_2=\frac{\sigma_2}{2\epsilon_0}=\frac{Q}{2A_2\epsilon_0},$$

Therefore, the integral becomes

$$\Delta V =\frac{Q}{2\epsilon_0}\frac{A_1 + A_2}{A_1A_2}d$$

in magnitude, and the capacitance is

$$ C=\frac{2\epsilon_0 A_1A_2}{d(A_1+A_2)}. $$

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  • $\begingroup$ It looks to me like your derivation assumes that the surface charge density is constant - which will not be true in the case where the two plates don't have the same area. This is trivial to see when one area is much smaller than the other, and the problem starts to look like a "point charge and infinite plane" - there will be an induced surface charge distribution to keep the E field perpendicular to the conductor. Does your derivation still work when you don't assume constant charge density? I don't see it. $\endgroup$ – Floris Jun 19 '17 at 15:08
  • $\begingroup$ I could be wrong, but I believe the issue with your point is in assuming that the number of charges remain constant as the area decreases. The issue I see with this is the fact that the charges in a conductor act to remove said conductor's internal electric field. That tends to mean that the charges end up evenly spaced inside the conductor. Not certain if this changes with the fact we're discussing a capacitor, where the electric field may be an issue, though. $\endgroup$ – DoublyNegative Feb 13 '18 at 18:46
  • $\begingroup$ I think inside a conductor charges move to reduce the tangential component of an electric field to zero. But the perpendicular (normal) electric field can vary if the surface density of charge varies (which I think would be the case in the plate with the larger area-the region opposite the smaller plate would have a higher charge density). $\endgroup$ – Kevin Jan 17 at 5:44
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Ok, I will try use latex.

Here is the charge distribution that I obtained for same area capacitor and the double radius relation ones:

Charge distribution of same sized plates and also with other with double radius one with respect the other

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  • $\begingroup$ Wow, nice graphs. They certainly show that the density in the big plate is higher in the region opposite the small plate. I wonder why the density increases at the outside edge of the big plate? $\endgroup$ – Kevin Jan 17 at 5:47
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I divided every plate in meshes and giving to one plate V=-0.5 and to another one V=0.5V. Then the voltage of every point can be calculated in the following way:

$$Vj=K*\sum \frac{Qi}{r_{ij}}$$ $$\begin{vmatrix} \\V0 \\V1 \\... \\Vn \end{vmatrix} =K*\begin{vmatrix} a &1/r01 &... &1/r0n \\ 1/r10 &a &... &1/r1n \\ ... &... &... &... \\ 1/rn0 &1/rn1 &... &1/rnn \end{vmatrix} *\begin{vmatrix} \\Q0 \\Q1 \\... \\Qn \end{vmatrix}$$ Where rij is the distance of the point i to the charge j And a=1/0.1*mesh size This can be written in an array form: [V]=K*[1/r]*[Q] So making the inverse of the array we have: $$[Q]=\frac{1}{K}*[1/r]^{-1}*[V]$$

He mesh size was 1mm. Then as long as C=Q/V we can calculate C in farads:

C= \frac{\sum Q}{V} Where V=0.5-0.5=1 volt I added absolute values of charges (if not, a large error is obtained)

In the calculus I had to make inverse of a 3000x3000 array using multithreads

Then obtained following results (SI units):

R1=0.02 R2=0.02 H=1e-3 capacitance=11.637pF (expected 11.13pF)

R1=0.01 R2=0.02 H=1e-3 capacitance=3.90355pF

The 3.9pF corresponds to a capacitor area for R=0.01185m that is just in the middle between the 2A1A2/(A1+A2) and the minor radius capacitance.

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  • $\begingroup$ Please use the $\LaTeX$ syntax to typeset equations, don't paste them as images or plain text. $\endgroup$ – Ruslan May 18 '18 at 12:12

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