,Solutions to Problem Sets zl zl zl
The selected solutions to all 12 chapters problem sets are presented in this manual. The problem sets
zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl
depict examples of practical applications of the concepts described in the book, more detailed anal
zl zl zl zl zl zl zl zl zl zl zl zl zl zl
ysis of some of the ideas, or in some cases present a new concept.
zl zl zl zl zl zl zl zl zl zl zl zl zl
Note that selected problems have been given answers already in the book.
zl zl zl zl zl zl zl zl zl zl zl
,1 Chapter One zl
1. Using spherical coordinates, find the capacitance formed by two concentric spherical con
zl zl zl zl zl zl zl zl zl zl zl
ducting shells of radius a, and b. What is the capacitance of a metallic marble with a diamet
zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl
er of 1cm in free space? Hint: let 𝑏 → ∞, thus, 𝐶 = 4𝜋𝜀𝜀0𝑎 = 0.55𝑝𝐹.
zl zl zl zl zl zl zl zl zl zl zl zl z l zl zl zl
Solution: Suppose the inner sphere has a surface charge density of +𝜌𝑆. The outer surface ch
zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl
arge density is negative, and proportionally smaller (by (𝑎/𝑏)2) to keep the total charge the sa
zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl
me.
-
+
+S - + a + -
b
+
-
From Gauss’s law:zl zl
ф𝐷 ⋅ 𝑑𝑆 = 𝑄𝑄 = +𝜌𝑆4𝜋𝑎2 zl zl zl zl zl zl
𝑆
Thus, inside the sphere (𝑎 ≤ 𝑟 ≤ 𝑏):
zl zl zl zl z l zl zl zl
𝑎2
𝐷 = 𝜌𝑆 2 𝑎𝑟
𝑟
zl zl zl
zl
Assuming a potential of 𝑉0 between the𝑎inner and
2 2 1 1
outer surfaces, we have:
𝜌 𝑎 𝑑𝑟 = 𝜌𝑆 𝑎
1zl zl zl zl zl zl zl z l zl z l zl zl zl zl
𝑉 =−
z l
zl
z l
z l zl zl z l zl z l
0 𝑆 ( − ) zl zlz l z l z
l
2
𝑏 𝑟 𝜖 𝑎 𝑏 z l z l
Thus: 𝜖
𝜌𝑆4𝜋𝑎2 = 4𝜋𝜖
𝑄𝑄 z l z l
𝐶 =𝑉 =
𝜌 𝑆 21 1 1 1 z l zl zl
z l
𝑎 −𝑏
z l
0
𝜖 𝑎 (𝑎 𝑏− ) zl
zl
zl
zl
1
In the case of a metallic marble, 𝑏 → ∞, and hence: 𝐶 = 4𝜋𝜀𝜀0 𝑎. Letting 𝜀𝜀0 =
zl zl zl zl zl zl zl zl zl zl zl zl z l zl zl zl
z l z l z l z l ×
36𝜋
5
10−9, and 𝑎 = 0.5𝑐𝑚, it yields 𝐶 = 𝑝𝐹 = 0.55𝑝𝐹.
zl
z l zl
zl zl z l zl zl zl z l z l z l
9
2. Consider the parallel plate capacitor containing two different dielectrics. Find the total ca
zl zl zl zl zl zl zl zl zl zl zl zl
pacitance as a function of the parameters shown in the figure.
zl zl zl zl zl zl zl zl zl zl
, Area: A zl
1
d1
2
d2
Solution: Since in the boundary no charge exists (perfect insulator), the normal component of t
zl zl zl zl zl zl zl zl zl zl zl zl zl zl
he electric flux density has to be equal in each dielectric. That is:
zl zl zl zl zl zl zl zl zl zl zl zl
𝐷1 = 𝐷𝟐𝟐 z l zl
Accordingly:
𝜖1𝐸1 = 𝜖2𝐸𝟐𝟐 z l z l
Assuming a surface charge density of +𝜌𝑆 for the top plate, and −𝜌𝑆 for the bottom plate, the el
zl zl zl zl zl zl z l zl zl zl zl zl z l zl zl zl zl zl
ectric field (or flux has a component only in z direction, and we have:
zl zl zl zl zl zl zl zl zl zl zl zl zl
𝐷1 = 𝐷𝟐𝟐 = −𝜌𝑆𝑎𝑧
zl zl zl zl
If the potential between the top ad bottom plates is 𝑉0, based on the line integral we obtain:
zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl
𝑑2 𝜌𝑆
𝑑1+𝑑2 zl
−𝜌𝑆 𝑑1+𝑑2 zl
−𝜌 𝜌𝑆
𝑉0 = − 𝐸. 𝑑𝑧 = − 𝜖 𝑑𝑧 − 𝑆 𝑑𝑧 = 𝜖 𝑑1 + 𝑑2 zl z l
zl zl zl zl zl zl zl zl
𝜖 𝜖
zl zl
z l z l
0 0 2 𝑑2 1 1 2
Since the total charge on each plate is: 𝑄𝑄 = 𝜌𝑆𝐴, the capacitance is found to be:
zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl
= 𝐴 zl
𝐶 = 𝑄𝑄 𝑑 𝑑
z l
z l z l
z l
𝑉
0 1 zl
2
+ 𝜖2 zl
𝜖1
which is analogous to two parallel capacitors.
zl zl zl zl zl zl
3. What would be the capacitance of the structure in problem 2 if there were a third conductor wi
zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl
th zero thickness at the interface of the dielectrics? How would the electric field lines look? H
zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl
ow does the capacitance change if the spacing between the top and bottom plates are kept the sa
zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl
me, but the conductor thickness is not zero?
zl zl zl zl zl zl zl
The selected solutions to all 12 chapters problem sets are presented in this manual. The problem sets
zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl
depict examples of practical applications of the concepts described in the book, more detailed anal
zl zl zl zl zl zl zl zl zl zl zl zl zl zl
ysis of some of the ideas, or in some cases present a new concept.
zl zl zl zl zl zl zl zl zl zl zl zl zl
Note that selected problems have been given answers already in the book.
zl zl zl zl zl zl zl zl zl zl zl
,1 Chapter One zl
1. Using spherical coordinates, find the capacitance formed by two concentric spherical con
zl zl zl zl zl zl zl zl zl zl zl
ducting shells of radius a, and b. What is the capacitance of a metallic marble with a diamet
zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl
er of 1cm in free space? Hint: let 𝑏 → ∞, thus, 𝐶 = 4𝜋𝜀𝜀0𝑎 = 0.55𝑝𝐹.
zl zl zl zl zl zl zl zl zl zl zl zl z l zl zl zl
Solution: Suppose the inner sphere has a surface charge density of +𝜌𝑆. The outer surface ch
zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl
arge density is negative, and proportionally smaller (by (𝑎/𝑏)2) to keep the total charge the sa
zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl
me.
-
+
+S - + a + -
b
+
-
From Gauss’s law:zl zl
ф𝐷 ⋅ 𝑑𝑆 = 𝑄𝑄 = +𝜌𝑆4𝜋𝑎2 zl zl zl zl zl zl
𝑆
Thus, inside the sphere (𝑎 ≤ 𝑟 ≤ 𝑏):
zl zl zl zl z l zl zl zl
𝑎2
𝐷 = 𝜌𝑆 2 𝑎𝑟
𝑟
zl zl zl
zl
Assuming a potential of 𝑉0 between the𝑎inner and
2 2 1 1
outer surfaces, we have:
𝜌 𝑎 𝑑𝑟 = 𝜌𝑆 𝑎
1zl zl zl zl zl zl zl z l zl z l zl zl zl zl
𝑉 =−
z l
zl
z l
z l zl zl z l zl z l
0 𝑆 ( − ) zl zlz l z l z
l
2
𝑏 𝑟 𝜖 𝑎 𝑏 z l z l
Thus: 𝜖
𝜌𝑆4𝜋𝑎2 = 4𝜋𝜖
𝑄𝑄 z l z l
𝐶 =𝑉 =
𝜌 𝑆 21 1 1 1 z l zl zl
z l
𝑎 −𝑏
z l
0
𝜖 𝑎 (𝑎 𝑏− ) zl
zl
zl
zl
1
In the case of a metallic marble, 𝑏 → ∞, and hence: 𝐶 = 4𝜋𝜀𝜀0 𝑎. Letting 𝜀𝜀0 =
zl zl zl zl zl zl zl zl zl zl zl zl z l zl zl zl
z l z l z l z l ×
36𝜋
5
10−9, and 𝑎 = 0.5𝑐𝑚, it yields 𝐶 = 𝑝𝐹 = 0.55𝑝𝐹.
zl
z l zl
zl zl z l zl zl zl z l z l z l
9
2. Consider the parallel plate capacitor containing two different dielectrics. Find the total ca
zl zl zl zl zl zl zl zl zl zl zl zl
pacitance as a function of the parameters shown in the figure.
zl zl zl zl zl zl zl zl zl zl
, Area: A zl
1
d1
2
d2
Solution: Since in the boundary no charge exists (perfect insulator), the normal component of t
zl zl zl zl zl zl zl zl zl zl zl zl zl zl
he electric flux density has to be equal in each dielectric. That is:
zl zl zl zl zl zl zl zl zl zl zl zl
𝐷1 = 𝐷𝟐𝟐 z l zl
Accordingly:
𝜖1𝐸1 = 𝜖2𝐸𝟐𝟐 z l z l
Assuming a surface charge density of +𝜌𝑆 for the top plate, and −𝜌𝑆 for the bottom plate, the el
zl zl zl zl zl zl z l zl zl zl zl zl z l zl zl zl zl zl
ectric field (or flux has a component only in z direction, and we have:
zl zl zl zl zl zl zl zl zl zl zl zl zl
𝐷1 = 𝐷𝟐𝟐 = −𝜌𝑆𝑎𝑧
zl zl zl zl
If the potential between the top ad bottom plates is 𝑉0, based on the line integral we obtain:
zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl
𝑑2 𝜌𝑆
𝑑1+𝑑2 zl
−𝜌𝑆 𝑑1+𝑑2 zl
−𝜌 𝜌𝑆
𝑉0 = − 𝐸. 𝑑𝑧 = − 𝜖 𝑑𝑧 − 𝑆 𝑑𝑧 = 𝜖 𝑑1 + 𝑑2 zl z l
zl zl zl zl zl zl zl zl
𝜖 𝜖
zl zl
z l z l
0 0 2 𝑑2 1 1 2
Since the total charge on each plate is: 𝑄𝑄 = 𝜌𝑆𝐴, the capacitance is found to be:
zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl
= 𝐴 zl
𝐶 = 𝑄𝑄 𝑑 𝑑
z l
z l z l
z l
𝑉
0 1 zl
2
+ 𝜖2 zl
𝜖1
which is analogous to two parallel capacitors.
zl zl zl zl zl zl
3. What would be the capacitance of the structure in problem 2 if there were a third conductor wi
zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl
th zero thickness at the interface of the dielectrics? How would the electric field lines look? H
zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl
ow does the capacitance change if the spacing between the top and bottom plates are kept the sa
zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl zl
me, but the conductor thickness is not zero?
zl zl zl zl zl zl zl
