# effective capacitance

Discussion in 'Electronic Design' started by [email protected], Feb 28, 2006.

1. ### Guest

What is the effective dielectric constant for a parallel-plate
constructed as follows:

| | |
| | |
-----| E1 | E2 |-----
| | |
| | |

where both section have the same thickness for simplicity. I don't
think you can averga E1 and E2, correct?

The reason I ask is because I am curious about what the effective
capacitance of a non-conducting pipe-wall would be if the measuring
electrode was "stood-off" from the pipe using a low-dielectric material
with the same thickness as the pipe (again, the same thickness assuming
here is for simplicity).

Thanks.

2. ### Tom BruhnsGuest

It's two capacitors in series. Imagine a thin conducting surface
between E1 and E2. For equal size plates on both ends, and assuming no
fringing effects around the edges, C1 is proportional to E1/t1 and C2
is proportional to E2/t2, where E1 and E2 are the permittivities (or
relative dielectric constants) and t1 and t2 are the plate spacings or
thincknesses. Since the net capacitance is C1*C2/(C1+C2), the net
capacitance is proportional to E1E2/(E1*t2+E2*t1). That can be
rewritten as 1/(t2/E2 + t1/E1). In other words, you can average the
1/E values, weighted by the thicknesses. (This also applies to the
case where the dielectrics are in a homogenous mixture, such as the
foam dielectrics commonly used in coaxial cables.)

Sanity check: if you have equal thicknesses and E1=999*E2, averaging
the permittivities yields 500*E2. But clearly the capacitance cannot
be more than that of just the E2 section alone.

Cheers,
Tom

3. ### Tim WescottGuest

If you do the ascii-art thing please use fixed-point font, and no tabs.

Your three-plate structure, with the middle plate floating, should
pretty much have the same capacitance as the left-to-middle capacitor in
series with the middle-to-right capacitor. You'll have some direct
coupling from left to right through fringe effects, but by and large you
should be able to estimate the capacitance.

I assume you don't really have a conductor in the middle, but are just
trying to break the problem down in approved engineering fashion. I
expect the answer will be substantially the same with or without that
conductor.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

4. ### Guest

Thanks. My worry was just that: the legitimacy of breaking down the
problem by assuming that conductor in the center. From the responses
I've seen here looks like it Ok. Thanks guys.

5. ### Guest

Thanks. My worry was just that: the legitimacy of breaking down the
problem by assuming that conductor in the center. From the responses
I've seen here looks like it Ok. Thanks guys.