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[Phys-L] Fw: non-polarized capacitor

From: John Denker <>
Sent: Tuesday, February 23, 2021 6:33 PM
To: Jeffrey Schnick
Subject: Re: [Phys-L] non-polarized capacitor

On 2/23/21 3:07 PM, you wrote:

"The floating node immediately picks up enough gorge to make it float
at a voltage" but by your definition of gorge, gorge is something
that a capacitor or a battery has, not something that a node has. Is
this just a typo?

I understand the question, and I understand why you ask.
We agree that it's unclear and confusing.

However I would argue that it's not wrong. Here's how I think
about it. Do you find any of these persuasive? Or does this
just bring additional confusion?

1) Certainly the node doesn't pick up any free charge Q.
So by process of elimination it has to be gorge.

2) Perhaps the simplest way to clarify it would be to cross
out what it says about gorge on the node, and instead say
that the two capacitors each pick up enough gorge to make
the floating node float at the required voltage.

During the initial transient, when the input voltage V is
high, the bottom capacitor leaks, putting gorge on the top
capacitor (but not the bottom), lowering Vc. When V is low,
the top capacitor leaks, putting gorge on the bottom capacitor
(but not the top), again lowering Vc.

3) I'm not sure I thought about it very clearly, but I think
the following complicated idea is what I had in mind:

The floating node has some capacitance to ground. Consider
the two (-) plates of the capacitors to be one side of a
double-sized capacitor, while the two (+) plates are the
other side. That is to say, the floating node sees two
capacitors in parallel. The input V is considered AC ground,
since we assume it is driven from a low-impedance source.

Leakage puts some gorge on this double-sized capacitor.

To make sense of this, you have to imagine each real capacitor
as an ideal capacitor plus some leakage. Beware that one
capacitor is leaking while the other is not, so the situation
is not symmetrical.