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parallel plates capacitor

Consider the following parallel plate capacitor

Givens:
top plate 1 meter wide by 1 meter deep by 10 mm thick
bottom plate 1 meter wide by 1 meter deep by 10 mm thick
gap 12 mm
applied voltage 1000 volts
containing different materials inside the gap as follows:

a) 12 mm of vacuum (Ke = 1)
b) 1 mm of vacuum, 10 mm of dielectric (Ke = 4), 1 mm of vacuum
c) 1 mm of vacuum, 10 mm of paraelectric (Ke = 20), 1 mm of vacuum
d) 1 mm of vacuum, 10 mm of ferroelectric (Ke = 300), 1 mm of vacuum
e) 0.3 mm vacuum, 3.8 mm dielectric, 3.8 mm paraelectric, 3.8 mm ferroelectric, 0.3 mm vacuum

CALCULATE each of the following for a) thru e) cases above:

1) Calculate E, D and P inside the vacuum gaps and inside each material
2) Calculate the resulting capacitance for each case
3) Calculate the force of electrical attraction when the material is in the center of the capacitor
4) Calculate the work required to pull the material out of the gap.

Solution Preview

The derivations and explanations are written in the attached pdf file.

Question (3) has ambiguity in it, however the combination of answers in (3) and (4) covers two possible interpretations. If you find among your other materials that the interpretation should be something else, you can re-post this question specifying the other interpretation and I shall help.

As regards the substitutions of the values of thicknesses and permittivities in various slabs of vacuum and materials, I have indicated at the start that the slabs of vacuum at the top and the bottom can be united into effective one slab for the purpose of calculations, and also gave some examples at the end of the answer to question (1). If you have hesitations about how to substitute in any particular cases, you can re-post these particulars and I shall help.

Question (4) is the good one, concerning a fine point of energy balance. I suggested looking up the relevant discussion in section 4.4.4 of Griffiths' book, but you may also have discussions of the same in your other sources.

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centerline{bf Parallel plate capacitor}

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bf Notations rm

Plate dimensions: $a_x=1~m$, $a_y=1~m$;
Area $A = a_x*a_y = 1~m^2$;
Gap $D=0.012~m$;
Applied constant voltage $V=1000~V$;
Vacuum thickness and relative permittivity in case (a):
$d_a=0.012~m$, $epsilon_{ra} = 1$;
Regular dielectric thickness and relative permittivity in case (b):
$d_{ba}=0.0001~m+0.0001~m=0.0002~m$,$d_{bb}=0.001~m$, $epsilon_{rb} = 4$;
Paraelectric thickness and relative permittivity in case (c):
$d_{ca}=0.0002~m$, $d_{cc}=0.001~m$, $epsilon_{rc} = 20$;
Ferroelectric thickness and ...

Solution Summary

Parallel plates capacitor is investigates, such as its capacitance and electric field.

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