Oops! Of course, Bob and Michael are right. And, of course, I
should have known there was something wrong with my derivation
because it is obvious (from superposition!) that one *can* add a
uniform background field to the field of the capacitor and obtain
any number of valid solutions like the ones below:
^ ^ ^ ^ | | | |
| | | | V V V V
+ + + + - - - -
conductor conductor
---------- ----------
^^^^^^^^^^ ^^^^^^^^^^
|||||||||| ||||||||||
++++++++++ ++++++++++
conductor conductor
- - - - + + + +
^ ^ ^ ^ | | | |
| | | | V V V V
On the other hand, it is still true that the *added* charge--the
charge that counts in V=Q/C *is* entirely on the inside surface.
If these capacitors were "fully discharged" they would look as
shown below and we would say that they "carry zero charge."
^ ^ ^ ^ | | | |
| | | | V V V V
+ + + + - - - -
conductor conductor
conductor conductor
- - - - + + + +
^ ^ ^ ^ | | | |
| | | | V V V V