,Solutions to Problem Sets BK BK BK
The selected solutions to all 12 chapters problem sets are presented in this manual. The pr
BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK
oblem sets depict examples of practical applications of the concepts described in the boo
BK BK BK BK BK BK BK BK BK BK BK BK BK
k, more detailed analysis of some of the ideas, or in some cases present a new concept.
BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK
Note that selected problems have been given answers already in the book.
BK BK BK BK BK BK BK BK BK BK BK
,1 Chapter One BK
1. Using spherical coordinates, find the capacitance formed by two concentric spher
BK BK BK BK BK BK BK BK BK BK
ical conducting shells of radius a, and b. What is the capacitance of a metallic m
BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK
arble with a diameter of 1cm in free space? Hint: let 𝑏 → ∞, thus, 𝐶 = 4𝜋𝜀𝜀0
BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK B K BK
𝑎 = 0.55𝑝𝐹.
BK BK
Solution: Suppose the inner sphere has a surface charge density of +𝜌𝑆. The outer
BK BK BK BK BK BK BK BK BK BK BK BK BK BK
surface charge density is negative, and proportionally smaller (by (𝑎/𝑏)2) to keep th
BK BK BK BK BK BK BK BK BK BK BK BK
e total charge the same.
BK BK BK BK
-
+
+S - + a + -
b
+
-
From Gauss’s law:
BK BK
ф𝐷 ⋅ 𝑑𝑆 = 𝑄𝑄 = +𝜌𝑆4𝜋𝑎2
BK BK BK BK BK BK
𝑆
Thus, inside the sphere (𝑎 ≤ 𝑟
BK BK BK BK BK BK BK
≤ 𝑏):
BK
𝑎2
𝐷 = 𝜌𝑆 2 𝑎𝑟 BK BK
𝑟 BK
BK
Assuming a potential of 𝑉0 between 𝑎 1the inner2 and 2 1 1
outer surfaces, we have:
𝜌 𝑎 𝑑𝑟 = 𝜌𝑆 𝑎
BK BK BK BK BK B K BK BK B K BK BK BK BK BK
𝑉 = − BK B K
BK
B K BK BK B K BK B K
0 𝑆 ( − ) BK BKB K B K B
K
2
𝑏 𝑟 𝜖 𝑎 𝑏 BKB
Thus: 𝜖 K
𝜌𝑆4𝜋𝑎2 = 4𝜋𝜖
𝑄𝑄 B K B K
𝐶 = 𝑉 =
𝜌 𝑆 21 1 1 1 BK BK BK
B K
B K
𝜖 𝑎 (𝑎 − 𝑏) 𝑎 − 𝑏
0 BK BK
BK BK
1
In the case of a metallic marble, 𝑏 → ∞, and hence: 𝐶 𝑎. Letting 𝜀 =
BK BK BK BK BK BK BK BK BK BK BK BK B BK BK
B K B K BKB
K= 4𝜋𝜀𝜀0
BK 𝜀0 × K
36𝜋
5
10−9, and 𝑎 = 0.5𝑐𝑚, it yields 𝐶 = 𝑝𝐹 = 0.55𝑝𝐹. BK
BK BK
BK BK BK BK BK BK B K BK B K
9
2. Consider the parallel plate capacitor containing two different dielectrics. Find the
BK BK BK BK BK BK BK BK BK BK B
total capacitance as a function of the parameters shown in the figure.
K BK BK BK BK BK BK BK BK BK BK BK
, Area: A BK
1
d1
2
d2
Solution: Since in the boundary no charge exists (perfect insulator), the normal compo
BK BK BK BK BK BK BK BK BK BK BK BK
nent of the electric flux density has to be equal in each dielectric. That is:
BK BK BK BK BK BK BK BK BK BK BK BK BK BK
𝐷1 = 𝐷𝟐𝟐 B K BK
Accordingly:
𝜖1𝐸1 = 𝜖2𝐸𝟐𝟐 B K BK
Assuming a surface charge density of +𝜌𝑆 for the top plate, and −𝜌𝑆 for the bottom
BK BK BK BK BK BK B K BK BK BK BK BK B K BK BK BK
plate, the electric field (or flux has a component only in z direction, and we have:
BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK
𝐷1 = 𝐷𝟐𝟐 = −𝜌𝑆𝑎𝑧
BK BK BK BK
If the potential between the top ad bottom plates is 𝑉0, based on the line integral we obtai
BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK
n:
𝑑1+𝑑2 𝑑2B K
−𝜌𝑆 𝑑1+𝑑2B K −𝜌 𝜌𝑆 𝜌𝑆
𝑉0 = − 𝐸. 𝑑𝑧 = − 𝑆 𝑑𝑧 = 𝑑1 + 𝑑2
𝜖 BK B K
BK BK BK BK BK BK BK
𝜖 𝜖 BK BK
𝜖
𝑑𝑧 BK
B K B K
−
0 0 2 𝑑2 1 1 2
Since the total charge on each plate is: 𝑄𝑄 = 𝜌𝑆𝐴, the capacitance is found to be:
BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK
= 𝐴 B K
BK
𝐶 = 𝑄𝑄 𝑑 𝑑 BK B K B K
𝑉
0 1B K 2
+ 𝜖 2
BK
𝜖1
which is analogous to two parallel capacit
BK BK BK BK BK BK
ors.
3. What would be the capacitance of the structure in problem 2 if there were a third c
BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK
onductor with zero thickness at the interface of the dielectrics? How would the electr
BK BK BK BK BK BK BK BK BK BK BK BK BK
ic field lines look? How does the capacitance change if the spacing between the top a
BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK
nd bottom plates are kept the same, but the conductor thickness is not zero?
BK BK BK BK BK BK BK BK BK BK BK BK BK
The selected solutions to all 12 chapters problem sets are presented in this manual. The pr
BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK
oblem sets depict examples of practical applications of the concepts described in the boo
BK BK BK BK BK BK BK BK BK BK BK BK BK
k, more detailed analysis of some of the ideas, or in some cases present a new concept.
BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK
Note that selected problems have been given answers already in the book.
BK BK BK BK BK BK BK BK BK BK BK
,1 Chapter One BK
1. Using spherical coordinates, find the capacitance formed by two concentric spher
BK BK BK BK BK BK BK BK BK BK
ical conducting shells of radius a, and b. What is the capacitance of a metallic m
BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK
arble with a diameter of 1cm in free space? Hint: let 𝑏 → ∞, thus, 𝐶 = 4𝜋𝜀𝜀0
BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK B K BK
𝑎 = 0.55𝑝𝐹.
BK BK
Solution: Suppose the inner sphere has a surface charge density of +𝜌𝑆. The outer
BK BK BK BK BK BK BK BK BK BK BK BK BK BK
surface charge density is negative, and proportionally smaller (by (𝑎/𝑏)2) to keep th
BK BK BK BK BK BK BK BK BK BK BK BK
e total charge the same.
BK BK BK BK
-
+
+S - + a + -
b
+
-
From Gauss’s law:
BK BK
ф𝐷 ⋅ 𝑑𝑆 = 𝑄𝑄 = +𝜌𝑆4𝜋𝑎2
BK BK BK BK BK BK
𝑆
Thus, inside the sphere (𝑎 ≤ 𝑟
BK BK BK BK BK BK BK
≤ 𝑏):
BK
𝑎2
𝐷 = 𝜌𝑆 2 𝑎𝑟 BK BK
𝑟 BK
BK
Assuming a potential of 𝑉0 between 𝑎 1the inner2 and 2 1 1
outer surfaces, we have:
𝜌 𝑎 𝑑𝑟 = 𝜌𝑆 𝑎
BK BK BK BK BK B K BK BK B K BK BK BK BK BK
𝑉 = − BK B K
BK
B K BK BK B K BK B K
0 𝑆 ( − ) BK BKB K B K B
K
2
𝑏 𝑟 𝜖 𝑎 𝑏 BKB
Thus: 𝜖 K
𝜌𝑆4𝜋𝑎2 = 4𝜋𝜖
𝑄𝑄 B K B K
𝐶 = 𝑉 =
𝜌 𝑆 21 1 1 1 BK BK BK
B K
B K
𝜖 𝑎 (𝑎 − 𝑏) 𝑎 − 𝑏
0 BK BK
BK BK
1
In the case of a metallic marble, 𝑏 → ∞, and hence: 𝐶 𝑎. Letting 𝜀 =
BK BK BK BK BK BK BK BK BK BK BK BK B BK BK
B K B K BKB
K= 4𝜋𝜀𝜀0
BK 𝜀0 × K
36𝜋
5
10−9, and 𝑎 = 0.5𝑐𝑚, it yields 𝐶 = 𝑝𝐹 = 0.55𝑝𝐹. BK
BK BK
BK BK BK BK BK BK B K BK B K
9
2. Consider the parallel plate capacitor containing two different dielectrics. Find the
BK BK BK BK BK BK BK BK BK BK B
total capacitance as a function of the parameters shown in the figure.
K BK BK BK BK BK BK BK BK BK BK BK
, Area: A BK
1
d1
2
d2
Solution: Since in the boundary no charge exists (perfect insulator), the normal compo
BK BK BK BK BK BK BK BK BK BK BK BK
nent of the electric flux density has to be equal in each dielectric. That is:
BK BK BK BK BK BK BK BK BK BK BK BK BK BK
𝐷1 = 𝐷𝟐𝟐 B K BK
Accordingly:
𝜖1𝐸1 = 𝜖2𝐸𝟐𝟐 B K BK
Assuming a surface charge density of +𝜌𝑆 for the top plate, and −𝜌𝑆 for the bottom
BK BK BK BK BK BK B K BK BK BK BK BK B K BK BK BK
plate, the electric field (or flux has a component only in z direction, and we have:
BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK
𝐷1 = 𝐷𝟐𝟐 = −𝜌𝑆𝑎𝑧
BK BK BK BK
If the potential between the top ad bottom plates is 𝑉0, based on the line integral we obtai
BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK
n:
𝑑1+𝑑2 𝑑2B K
−𝜌𝑆 𝑑1+𝑑2B K −𝜌 𝜌𝑆 𝜌𝑆
𝑉0 = − 𝐸. 𝑑𝑧 = − 𝑆 𝑑𝑧 = 𝑑1 + 𝑑2
𝜖 BK B K
BK BK BK BK BK BK BK
𝜖 𝜖 BK BK
𝜖
𝑑𝑧 BK
B K B K
−
0 0 2 𝑑2 1 1 2
Since the total charge on each plate is: 𝑄𝑄 = 𝜌𝑆𝐴, the capacitance is found to be:
BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK
= 𝐴 B K
BK
𝐶 = 𝑄𝑄 𝑑 𝑑 BK B K B K
𝑉
0 1B K 2
+ 𝜖 2
BK
𝜖1
which is analogous to two parallel capacit
BK BK BK BK BK BK
ors.
3. What would be the capacitance of the structure in problem 2 if there were a third c
BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK
onductor with zero thickness at the interface of the dielectrics? How would the electr
BK BK BK BK BK BK BK BK BK BK BK BK BK
ic field lines look? How does the capacitance change if the spacing between the top a
BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK
nd bottom plates are kept the same, but the conductor thickness is not zero?
BK BK BK BK BK BK BK BK BK BK BK BK BK
