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,Solutions to Problem Sets FC FC FC




The selected solutions to all 12 chapters problem sets are presented in this manual. The prob
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lem sets depict examples of practical applications of the concepts described in the book, m
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ore detailed analysis of some of the ideas, or in some cases present a new concept.
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Note that selected problems have been given answers already in the book.
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,1 Chapter One FC



1. Using spherical coordinates, find the capacitance formed by two concentric spheric
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al conducting shells of radius a, and b. What is the capacitance of a metallic marbl
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e with a diameter of 1cm in free space? Hint: let 𝑏 → ∞, thus, 𝐶 = 4𝜋𝜀𝜀0𝑎 = 0.
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55𝑝𝐹.

Solution: Suppose the inner sphere has a surface charge density of +𝜌𝑆. The outer su
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rface charge density is negative, and proportionally smaller (by (𝑎/𝑏)2) to keep the to
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tal charge the same.
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-
+

+S - + a + -

b
+
-

From Gauss’s law:FC FC




ф𝐷 ⋅ 𝑑𝑆 = 𝑄𝑄 = +𝜌𝑆4𝜋𝑎2
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𝑆
Thus, inside the sphere (𝑎 ≤ 𝑟 ≤
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𝑏):
FC
𝑎2
𝐷 = 𝜌𝑆 2 𝑎𝑟 FC FC
𝑟 FC
FC


Assuming a potential of 𝑉0 between 𝑎 1the inner
2 and 2 1 1 we have:
outer surfaces,
𝜌 𝑎 𝑑𝑟 = 𝜌𝑆 𝑎
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𝑉 = − FC F C
FC
F C FC FC F C FC F C


0 𝑆 ( − ) FC FCF C F C F
C

2
𝑏 𝑟 𝜖 𝑎 𝑏 F C F



Thus: 𝜖 C

𝜌𝑆4𝜋𝑎2 = 4𝜋𝜖
𝑄𝑄 F C F C



𝐶 =𝑉 =
𝜌 𝑆 21 1 1 1 F C FC FC
F C
F C


𝜖 𝑎 (𝑎 − 𝑏) 𝑎 − 𝑏
0 FC FC
FC FC

1
In the case of a metallic marble, 𝑏 → ∞, and hence: 𝐶 = 𝑎. Letting 𝜀 =
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F C F C FCFC



4𝜋𝜀𝜀0
FC 𝜀0 ×
36𝜋
5
10−9, and 𝑎 = 0.5𝑐𝑚, it yields 𝐶 = 𝑝𝐹 = 0.55𝑝𝐹. FC
F C FC
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9




2. Consider the parallel plate capacitor containing two different dielectrics. Find the t
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otal capacitance as a function of the parameters shown in the figure.
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, Area: A FC




1




d1
2




d2
Solution: Since in the boundary no charge exists (perfect insulator), the normal compon
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ent of the electric flux density has to be equal in each dielectric. That is:
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𝐷1 = 𝐷𝟐𝟐
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Accordingly:

𝜖1𝐸1 = 𝜖2𝐸𝟐𝟐 F C FC




Assuming a surface charge density of +𝜌𝑆 for the top plate, and −𝜌𝑆 for the bottom pl
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ate, the electric field (or flux has a component only in z direction, and we have:
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𝐷1 = 𝐷𝟐𝟐 = −𝜌𝑆𝑎𝑧
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If the potential between the top ad bottom plates is 𝑉0, based on the line integral we obtain:
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𝑑1+𝑑2 𝑑2F C
−𝜌𝑆 𝑑1+𝑑2F C −𝜌 𝜌𝑆 𝜌𝑆
𝑉0 = − 𝐸. 𝑑𝑧 = − 𝑆 𝑑𝑧 = 𝑑1 + 𝑑2
𝜖 FC F C
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𝜖 𝜖 FC FC
𝜖
𝑑𝑧 FC
F C F C



−
0 0 2 𝑑2 1 1 2

Since the total charge on each plate is: 𝑄𝑄 = 𝜌𝑆𝐴, the capacitance is found to be:
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= 𝐴 F C
FC


𝐶 = 𝑄𝑄 𝑑 𝑑 F C F C F C


𝑉
0 1F C 2
+ 𝜖2 FC

𝜖1
which is analogous to two parallel capacito
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rs.



3. What would be the capacitance of the structure in problem 2 if there were a third con
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ductor with zero thickness at the interface of the dielectrics? How would the electric fi
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eld lines look? How does the capacitance change if the spacing between the top and bot
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tom plates are kept the same, but the conductor thickness is not zero?
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