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From: Wanshan Li on 17 Oct 2007 16:59
Suppose that a parallel plate capacitor has rectangular plates but the
plates are not exactly parallel. The separation at one edge is d-a and d+a
at the other edge. (a<<d) Show that the capacitance is given approximately
by

C=(eo)*A/d*[1+a^2/(3d^2)].

where A is the area of a plate and eo is the permittivity of free space.

Thanks.

Wanshan Li


From: Helmut Wabnig hwabnig on 18 Oct 2007 03:47
On Wed, 17 Oct 2007 13:59:41 -0700, "Wanshan Li" <wanshan(a)ispwest.com>
wrote:

>Suppose that a parallel plate capacitor has rectangular plates but the
>plates are not exactly parallel. The separation at one edge is d-a and d+a
>at the other edge. (a<<d) Show that the capacitance is given approximately
>by
>
> C=(eo)*A/d*[1+a^2/(3d^2)].
>
>where A is the area of a plate and eo is the permittivity of free space.
>
>Thanks.
>
>Wanshan Li
>
because it is inverse dependent from d there is a logarithmic
function integral.

integral (1/x) is a logarithm.

When developing the series for the log
and taking the first element only we end up with a certain factor...


http://retina.et.tudelft.nl/data/artwork/publication/mems/743862331.pdf
From: PD on 18 Oct 2007 09:26
On Oct 17, 3:59 pm, "Wanshan Li" <wans...(a)ispwest.com> wrote:
> Suppose that a parallel plate capacitor has rectangular plates but the
> plates are not exactly parallel. The separation at one edge is d-a and d+a
> at the other edge. (a<<d) Show that the capacitance is given approximately
> by
>
> C=(eo)*A/d*[1+a^2/(3d^2)].
>
> where A is the area of a plate and eo is the permittivity of free space.
>
> Thanks.
>
> Wanshan Li

OK, here's an unusual path that may be of help. You probably know that
the energy that's stored in a capacitor is related to the capacitance
and the voltage applied to the plates. You probably also know there's
a relationship between the energy stored in an electric field and the
value of the field. So if you can find the energy stored in the
electric field for a particular voltage, then you can find the
capacitance, by putting these two relations together.

Here you have a case where, because a<<d, then the field is not too
distorted (curved) from the parallel plate configuration, so you can
still consider the field to be straight across from plate to plate for
the purposes of this calculation. But it varies in strength across the
face of the plate, being stronger at the d-a end and weaker at the d+a
end.

Sounds like a straightforward 1-D integral to me.

PD

From: PD on 18 Oct 2007 11:22
On Oct 18, 8:26 am, PD <TheDraperFam...(a)gmail.com> wrote:
> On Oct 17, 3:59 pm, "Wanshan Li" <wans...(a)ispwest.com> wrote:
>
> > Suppose that a parallel plate capacitor has rectangular plates but the
> > plates are not exactly parallel. The separation at one edge is d-a and d+a
> > at the other edge. (a<<d) Show that the capacitance is given approximately
> > by
>
> > C=(eo)*A/d*[1+a^2/(3d^2)].
>
> > where A is the area of a plate and eo is the permittivity of free space.
>
> > Thanks.
>
> > Wanshan Li
>
> OK, here's an unusual path that may be of help. You probably know that
> the energy that's stored in a capacitor is related to the capacitance
> and the voltage applied to the plates. You probably also know there's
> a relationship between the energy stored in an electric field and the
> value of the field. So if you can find the energy stored in the
> electric field for a particular voltage, then you can find the
> capacitance, by putting these two relations together.
>
> Here you have a case where, because a<<d, then the field is not too
> distorted (curved) from the parallel plate configuration, so you can
> still consider the field to be straight across from plate to plate for
> the purposes of this calculation. But it varies in strength across the
> face of the plate, being stronger at the d-a end and weaker at the d+a
> end.
>
> Sounds like a straightforward 1-D integral to me.
>
> PD

If this didn't help, then do this: Do what I suggest but for a
parallel plate capacitor. That is, calculate the electric field
between the plates of a capacitor, as a function of V and d. (This
should be easy.) Then calculate the energy density stored in that
electric field. Now integrate that energy density across the volume
between the plates of the capacitor. This is the total energy stored
in that capacitor and you should now use the relationship between
capacitance and stored energy to find the capacitance. If it all works
out right, you'll find a familiar expression for the capacitance.

Now do it again, but here the field (and so the energy density) will
be a function of a coordinate stretching from the narrow end of the
gap to the wide end of the gap.

PD

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