Electric fields change in the spacing between the plates?

[tsgkl]

Homework Statement

A,B,C,D are four 'thin',similar metallic parallel plates,equally separated by distance d,and connected to a cell of potential difference(V) as shown in the attachement.

1. Write the potentials of A,B,C,D.
2. If B and C be connected by a wire,then what will be the potentials of the plates?
3. How will the electric fields change in the spacing between the plates?
4. Will the charges on the plates A and D change?

Homework Equations

C=\frac{εA}{d}
(for parallel plate capacitor with air as dielectric)
V=\frac{E}{d}

The Attempt at a Solution

for part(1) i know that potential for A is V and for D is 0 as it is earthed,but i am a bit confused for B and C
for part(2) for B and C, V=1/2[2V/3+V/3]=V/2. for A and D, V is same as in (1)
for part(3) the electric field between A and B will increase from V/3d to V/2d,become 0 between B and C and increase from V/3d to V/2d between C and D.
i have little clue on part(4)


Please clear my confusions...

 

[gneill]

For part (1) you can use the geometry of the situation to determine the potential w.r.t. plate D given that the field should be uniform between A and D (plates B and C have a neutral net charge and are supposedly "thin", so they won't unduly disturb the overall field even if charge separation occurs between surfaces).

Alternatively, if you want you can look at the setup as an electronic circuit where adjacent pairs of plates form individual capacitors. In the initial setup there are effectively three equal capacitors in series. Work out the potentials using circuit rules.

It looks like you did okay for part (2); presumably you took advantage of the geometry and the fact that the wire between B and C forces them to be at the same potential.

For part (4), if you consider the electronic circuit point of view you've gone from three capacitors in series to only two. How might this effect the net capacitance as "seen" by the voltage supply?