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Re: A funny capacitor.



William Beaty wrote:

.... I see that bringing a conductive object near
the gap between the two plates tends to replace
"empty space" with two new capacitors in series, ...

Suppose it is correct to view the "funny capacitor"
as a combination of two capacitors in series. [I do
not think this this correct for a system in which
some lines of E go directly from one vertical plate
to another]. Under this [unrealistic] assumption,
and accounting for the left-right symmetry, we have:

1/C2 = 1/C3 + 1/C3 = 2/C3 and C2=C3/2

where C3 is the capacity of each component. The
prediction that C2>C1 would be correct only if
C3 were larger than 2*C1. How can one be sure that
this is true?

We know that C1 refers to the capacitance between the
two plates in setup A (when plate 3 is absent). But
what is C3? It is C of the capacitor between the
undefined part of a vertical plate and one half of
the horizontal plate in setup B. It is not at all
obvious to me that C3 is larger than 2*C1. It would
probably be useful to show that the setup B is
equivalent to a network of several capacitors
connected in parallel and in series. Knowing the
values of components we would be able to compare C2
with C1. But is this possible? I do not think so.
Ludwik Kowalski