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.