AL phy 2003 Q.26

Since the plates are isolated, i.e do not connect to any battery, the charges on each of them are constant.

The electric field arising from the plates thus remains unchanged.
By Gauss's Law,
electric field intensity = (surface charge density on each plate)/(permittivity of free space)

Because potential of plate B = electric field intensity x plate separation
the potential of plate B thus decreases when the separation between the plates reduces, as the electric field is constant throughout.

2011-04-17 22:34:24 補充：
The two-plate arrangement forms a parallel-plate capacitor. Capacitance C is given by
C = Q/V, where Q is the charge on each plate, V is the potential of plate B. ...

2011-04-17 22:35:46 補充：
Reducing the separation means increasing the capacitance. Because Q is kept unchanged (isolated plates), V would be decreases.

2011-04-17 22:38:49 補充：
Now, electric field intensity E = V/d, where d is the palte separation, reducing d leads to reducing V, and because the filed is a uniform field, hence E remains constant.

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