Purcell Tricky Three Plate Capacitor Questions

[Maxophonist]

This is problem 3.8 from Purcell's E&M.

Three conducting plates are placed parallel to one another. The outer plates are connected by a wire. The inner plate is isolated and carries a charge of 10 esu per cm^2. In what proportion must this charge divide itself into a surface charge on the top face and a surface charge on the bottom face of the middle plate.

Top plate
-------------------------------------
-------------------------------------
^
5cm
v
Middle plate
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------------------------------------

^
8cm
v
Bottom plate
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-----------------------------------

Applying the limit of large or even infinite area, my approach is based on the idea that the two connected plates must be at the same potential and in particular have the same potential difference from the middle plate...
QT / CT = QB / CB ==> QT 4 pi sT / A = QB 4 pi sB / A == >

sigT sT = sigB sB

Where sig is the surface charge on the one side of the plate and s is the distance between the middle plate and one of the others. Combine with

sigT + sigB = sigTot = 10 esu/cm^2

to conclude:

sigB = sigTot (sT/(sT+sB)) = 50/13 esu/cm^2
sigT = sigTot (sB/(sT+sB)) = 80/13 esu/cm^2

It follows that the inner surfaces of the outer plates must have the negative values of these charges.

What I am struggling with is how the charge is distributed on the outer surfaces of the outer plates. For example if the outer plate configuration has no net charge then are the outer surfaces the opposite of the inner surfaces (80/13 and 50/15 again) or are they equal (5 and 5)? What if the outer configuration starts with a charge Q?

Finally, however the charges are distributed can the six charged surfaces (top and bottom of each plane) be treated as six infinite sheets of charge? If so doesn't this give some residual field inside the conductors?

 

[NateD]

The answer to your questions are as follows: if the outer plate configuration has no net charge, then the outer surfaces will be the opposite of the inner surfaces (80/13 and 50/15). If the outer configuration starts with a charge Q, then the outer surfaces will have charges proportional to the initial charge.Yes, you can treat the six charged surfaces as six infinite sheets of charge. This will result in a residual field inside the conductors, which is calculated by taking the divergence of the electric field at each point.