,Solutions to Problem Sets NJ NJ NJ
The selected solutions to all 12 chapters problem sets are presented in this manual. The probl
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
em sets depict examples of practical applications of the concepts described in the book, mor
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
e detailed analysis of some of the ideas, or in some cases present a new concept.
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
Note that selected problems have been given answers already in the book.
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
,1 Chapter One NJ
1. Using spherical coordinates, find the capacitance formed by two concentric spherica
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
l conducting shells of radius a, and b. What is the capacitance of a metallic marble
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
with a diameter of 1cm in free space? Hint: let 𝑏 → ∞, thus, 𝐶 = 4𝜋𝜀𝜀0𝑎 = 0.55𝑝
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ N J NJ NJ NJ
𝐹.
Solution: Suppose the inner sphere has a surface charge density of +𝜌𝑆. The outer sur
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
face charge density is negative, and proportionally smaller (by (𝑎/𝑏)2) to keep the total
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
charge the same.
NJ NJ NJ
-
+
+S - + a + -
b
+
-
From Gauss’s law: NJ NJ
ф𝐷 ⋅ 𝑑𝑆 = 𝑄𝑄 = +𝜌𝑆4𝜋𝑎2
NJ NJ NJ NJ NJ NJ
𝑆
Thus, inside the sphere (𝑎 ≤ 𝑟 ≤
NJ NJ NJ NJ NJ NJ NJ N
𝑏):
J
𝑎2
𝐷 = 𝜌𝑆 2 𝑎𝑟 NJ NJ
𝑟 NJ
NJ
Assuming a potential of 𝑉0 between 𝑎 1the inner
2 and 2 1 1we have:
outer surfaces,
𝜌 𝑎 𝑑𝑟 = 𝜌𝑆 𝑎
NJ NJ NJ NJ NJ N JNJ NJ N J NJ NJ NJ NJ NJ
𝑉 = − NJ N J
NJ
N J NJ NJ N J NJ N J
0 𝑆 ( − ) NJ NJN J N J N
J
2
𝑏 𝑟 𝜖 𝑎 𝑏 N JN
Thus: 𝜖 J
𝜌𝑆4𝜋𝑎2 = 4𝜋𝜖
𝑄𝑄 N J N J
𝐶 =𝑉 =
𝜌 𝑆 21 1 1 1 N J NJ NJ
N J
N J
𝜖 𝑎 (𝑎 − 𝑏) 𝑎 − 𝑏
0 NJ NJ
NJ NJ
1
In the case of a metallic marble, 𝑏 → ∞, and hence: 𝐶 = 𝑎. Letting 𝜀𝜀 =
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
N J N J NJNJ
4𝜋𝜀𝜀0 0
×
36𝜋
5
10−9, and 𝑎 = 0.5𝑐𝑚, it yields 𝐶 = 𝑝𝐹 = 0.55𝑝𝐹.
NJ
N J NJ
NJ NJ NJ NJ NJ NJ N J NJ N J
9
2. Consider the parallel plate capacitor containing two different dielectrics. Find the to
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
tal capacitance as a function of the parameters shown in the figure.
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
, Area: A NJ
1
d1
2
d2
Solution: Since in the boundary no charge exists (perfect insulator), the normal compone
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
nt of the electric flux density has to be equal in each dielectric. That is:
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
𝐷1 = 𝐷𝟐𝟐N J NJ
Accordingly:
𝜖1𝐸1 = 𝜖2𝐸𝟐𝟐 N J NJ
Assuming a surface charge density of +𝜌𝑆 for the top plate, and −𝜌𝑆 for the bottom plat
NJ NJ NJ NJ NJ NJ N J NJ NJ NJ NJ NJ N J NJ NJ NJ
e, the electric field (or flux has a component only in z direction, and we have:
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
𝐷1 = 𝐷𝟐𝟐 = −𝜌𝑆𝑎𝑧
NJ NJ NJ NJ
If the potential between the top ad bottom plates is 𝑉0, based on the line integral we obtain:
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
𝑑1+𝑑2 𝑑2N J
−𝜌𝑆 𝑑1+𝑑2N J −𝜌 𝜌𝑆 𝜌𝑆
𝑉0 = − 𝐸. 𝑑𝑧 = − 𝑆 𝑑𝑧 = 𝑑1 + 𝑑2
𝜖 NJ N J
NJ NJ NJ NJ NJ NJ NJ
𝜖 𝜖 NJ NJ
𝜖
𝑑𝑧 NJ
N J N J
−
0 0 2 𝑑2 1 1 2
Since the total charge on each plate is: 𝑄𝑄 = 𝜌𝑆𝐴, the capacitance is found to be:
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
= 𝐴 N J
NJ
𝐶 = 𝑄𝑄 𝑑 𝑑 N J N J N J
𝑉
0 1N J 2
+ 𝜖2 NJ
𝜖1
which is analogous to two parallel capacitor
NJ NJ NJ NJ NJ NJ
s.
3. What would be the capacitance of the structure in problem 2 if there were a third cond
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
uctor with zero thickness at the interface of the dielectrics? How would the electric fiel
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
d lines look? How does the capacitance change if the spacing between the top and botto
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
m plates are kept the same, but the conductor thickness is not zero?
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
The selected solutions to all 12 chapters problem sets are presented in this manual. The probl
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
em sets depict examples of practical applications of the concepts described in the book, mor
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
e detailed analysis of some of the ideas, or in some cases present a new concept.
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
Note that selected problems have been given answers already in the book.
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
,1 Chapter One NJ
1. Using spherical coordinates, find the capacitance formed by two concentric spherica
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
l conducting shells of radius a, and b. What is the capacitance of a metallic marble
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
with a diameter of 1cm in free space? Hint: let 𝑏 → ∞, thus, 𝐶 = 4𝜋𝜀𝜀0𝑎 = 0.55𝑝
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ N J NJ NJ NJ
𝐹.
Solution: Suppose the inner sphere has a surface charge density of +𝜌𝑆. The outer sur
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
face charge density is negative, and proportionally smaller (by (𝑎/𝑏)2) to keep the total
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
charge the same.
NJ NJ NJ
-
+
+S - + a + -
b
+
-
From Gauss’s law: NJ NJ
ф𝐷 ⋅ 𝑑𝑆 = 𝑄𝑄 = +𝜌𝑆4𝜋𝑎2
NJ NJ NJ NJ NJ NJ
𝑆
Thus, inside the sphere (𝑎 ≤ 𝑟 ≤
NJ NJ NJ NJ NJ NJ NJ N
𝑏):
J
𝑎2
𝐷 = 𝜌𝑆 2 𝑎𝑟 NJ NJ
𝑟 NJ
NJ
Assuming a potential of 𝑉0 between 𝑎 1the inner
2 and 2 1 1we have:
outer surfaces,
𝜌 𝑎 𝑑𝑟 = 𝜌𝑆 𝑎
NJ NJ NJ NJ NJ N JNJ NJ N J NJ NJ NJ NJ NJ
𝑉 = − NJ N J
NJ
N J NJ NJ N J NJ N J
0 𝑆 ( − ) NJ NJN J N J N
J
2
𝑏 𝑟 𝜖 𝑎 𝑏 N JN
Thus: 𝜖 J
𝜌𝑆4𝜋𝑎2 = 4𝜋𝜖
𝑄𝑄 N J N J
𝐶 =𝑉 =
𝜌 𝑆 21 1 1 1 N J NJ NJ
N J
N J
𝜖 𝑎 (𝑎 − 𝑏) 𝑎 − 𝑏
0 NJ NJ
NJ NJ
1
In the case of a metallic marble, 𝑏 → ∞, and hence: 𝐶 = 𝑎. Letting 𝜀𝜀 =
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
N J N J NJNJ
4𝜋𝜀𝜀0 0
×
36𝜋
5
10−9, and 𝑎 = 0.5𝑐𝑚, it yields 𝐶 = 𝑝𝐹 = 0.55𝑝𝐹.
NJ
N J NJ
NJ NJ NJ NJ NJ NJ N J NJ N J
9
2. Consider the parallel plate capacitor containing two different dielectrics. Find the to
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
tal capacitance as a function of the parameters shown in the figure.
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
, Area: A NJ
1
d1
2
d2
Solution: Since in the boundary no charge exists (perfect insulator), the normal compone
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
nt of the electric flux density has to be equal in each dielectric. That is:
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
𝐷1 = 𝐷𝟐𝟐N J NJ
Accordingly:
𝜖1𝐸1 = 𝜖2𝐸𝟐𝟐 N J NJ
Assuming a surface charge density of +𝜌𝑆 for the top plate, and −𝜌𝑆 for the bottom plat
NJ NJ NJ NJ NJ NJ N J NJ NJ NJ NJ NJ N J NJ NJ NJ
e, the electric field (or flux has a component only in z direction, and we have:
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
𝐷1 = 𝐷𝟐𝟐 = −𝜌𝑆𝑎𝑧
NJ NJ NJ NJ
If the potential between the top ad bottom plates is 𝑉0, based on the line integral we obtain:
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
𝑑1+𝑑2 𝑑2N J
−𝜌𝑆 𝑑1+𝑑2N J −𝜌 𝜌𝑆 𝜌𝑆
𝑉0 = − 𝐸. 𝑑𝑧 = − 𝑆 𝑑𝑧 = 𝑑1 + 𝑑2
𝜖 NJ N J
NJ NJ NJ NJ NJ NJ NJ
𝜖 𝜖 NJ NJ
𝜖
𝑑𝑧 NJ
N J N J
−
0 0 2 𝑑2 1 1 2
Since the total charge on each plate is: 𝑄𝑄 = 𝜌𝑆𝐴, the capacitance is found to be:
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
= 𝐴 N J
NJ
𝐶 = 𝑄𝑄 𝑑 𝑑 N J N J N J
𝑉
0 1N J 2
+ 𝜖2 NJ
𝜖1
which is analogous to two parallel capacitor
NJ NJ NJ NJ NJ NJ
s.
3. What would be the capacitance of the structure in problem 2 if there were a third cond
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
uctor with zero thickness at the interface of the dielectrics? How would the electric fiel
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
d lines look? How does the capacitance change if the spacing between the top and botto
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
m plates are kept the same, but the conductor thickness is not zero?
NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ NJ
