Re: cheap way to test "doorknob" capacitors? (fwd)

[<sroys@xxxxxxxxxxxxxxxxx>]

---------- Forwarded message ----------
Date: Thu, 11 Nov 2004 14:40:18 -0600
From: Shaun Epp <scepp@xxxxxxx>
To: High Voltage list <hvlist@xxxxxxxxxx>
Subject: Re: cheap way to test "doorknob" capacitors? (fwd)


----- Original Message ----- 
From: "High Voltage list" <hvlist@xxxxxxxxxx>
To: "hvlist" <hvlist@xxxxxxxxxxxxxxxxx>
Sent: Thursday, November 11, 2004 9:27 AM
Subject: Re: cheap way to test "doorknob" capacitors? (fwd)


> Original poster: <sroys@xxxxxxxxxxxxxxxxx>
>
>
>
> ---------- Forwarded message ----------
> Date: Thu, 11 Nov 2004 07:09:50 -0700
> From: Gomez Addams <gomez@xxxxxxxxxxxx>
> To: High Voltage list <hvlist@xxxxxxxxxx>
> Subject: Re: cheap way to test "doorknob" capacitors? (fwd)
>
>
> On Nov 10, 2004, at 7:14 PM, High Voltage list wrote:
>
>> Original poster: <sroys@xxxxxxxxxxxxxxxxx>
>>
>>
>>
>> ---------- Forwarded message ----------
>> Date: Wed, 10 Nov 2004 19:31:13 -0600
>> From: Shaun Epp <scepp@xxxxxxx>
>> To: High Voltage list <hvlist@xxxxxxxxxx>
>> Subject: Re: cheap way to test "doorknob" capacitors? (fwd)
>>
>> I read that the capacitance of door knob caps are a function of
>> voltage.
>> The capacitance is higher at high voltages.
>
> Wouldn't be a very good capacitor if that were true.  It's news to me.
> Got a reference?
>

Unfortunatly I don't have a reference, but I remember the explanation saying 
it had to do with the electric field moving further outwards from the plates 
icreasing the effective capacitance as the voltage across the plates 
increases.

Now that I think of it, this doesn't sound right.  The electric field will 
always be present around the plates with voltage applied, more voltage would 
cause a stronger field.  Now I'm not sure!.  The only way this could happen 
is if the dielectric was a semiconductor and the voltage cause partial 
conduction of the dielectric effectively moving the plates closer together. 
Which doesn't sound right either.

Oh well,

shaun