Re: electrostatic confusion

From: John O'Flaherty (quiasmox_at_yahoo.com)
Date: 02/27/05 

Date: Sun, 27 Feb 2005 08:14:43 -0600

CWatters wrote:
> "John O'Flaherty" <quiasmox@yahoo.com> wrote in message
> news:38dcaiF5m0v24U1@individual.net...
>
>>>But the book gives 2.5 kN/m as the answer to (b). Can anyone show me
>>
>>where I'm going wrong?
>
>
> See if this thread helps...
>
> http://www.edaboard.com/viewtopic.php?p=404849
>
> If I understand it correctly...
>
> The force on a charge in a uniform field is F=EQ however in your case the
> field isn't uniform it's on "one side" of the charge so F = E*Q/2
>
> Sorry if that's not a very satisfactory explanation.

  It's helping a little. In the problem I had, the electric field
between the plates is unambiguous, being determined by the voltage and
the distance. Now, the discussion you pointed to suggests that only half
the charge on each plate feels the electric field. The discussion seems
to say that the surface charge on the inner surface of the plates is
only half in the field, since there is a field gradient between the
inside of an infinitesimal layer on the surface containing the charge,
to the outside of that surface.
  Is that the explanation, or is it that half the charge resides on the
backs of the plates and so is external to the field? Or is all the
charge attracted to the inner surface of the plates? 

--
john

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