A parallel plate capacitor with a nonuniform dielectric
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A parallel plate capacitor with a nonuniform dielectric
The dielectric of a parallel plate capacitor has a permittivity that varies as ero+ax, where x is the distance from one plate. The area of a plate is A, and their spacing is s.
(a) Find the capacitance.
(b) Show that, if er varies from ero to 2ero, then c is 1.44 times as large as if a were zero.
(c) Find P from the values of D and E for that case.
(d) Deduce the value of rb.
(e) Now calculate rb from the relation given in problem 7-2
[Prob 7-2: show that in a non homogeneous dielectric, if rf = 0, then
rb = (ero/er) E.grad er]
(f) Draw curves of e, rb and P as functions of x for ero = 3.00, a = ero/s, s = 1.00 mm when the applied voltage is 1.00 volts.
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Solution Summary
This problem deals with a parallel plate capacitor that has a dielectric with non uniform permittivity. There are several parts to this question. Quantities such as capacitance, polarization, displacement vector, electric field and bound charge density have been found. Answer is in a 6-page word document. Please download this challenging solution set, if you are looking to improve your problem solving skills in this field.
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Please see the attached file.
(a) Consider an element of thickness dx within the capacitor.
In general, capacitance can be written as C = A ε / s.
Using this expression, capacitance of the small element,
dC = A εo εr(x) / dx
Entire capacitor can be considered to be made up of several such elements connected in series. Using the fact that the resultant capacitance of capacitors connected in series is given by 1/C = 1/C1 + 1/C2 + ....we can write the following:
1/C = ∫ 1/ dC { limit of this integration is x from 0 to s)
1/C = ∫ 1/ dC = ∫ dx/ A εo εr(x) = ∫ dx/ A εo (εro + ax)
= (1/aAεo) ln (εro + ax) limit of this integration is x from 0 to s)
= (1/aAεo) ln [(εro + as)/ εro]
C = aAεo / ln [1 + as/εro]
(b) If εr = (εro + ax) varies of εro to 2 εro
at x = s, εr = 2 εro
substituting this on the expression εr = ...
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