,Solutions to Problem Sets gy gy gy
The selected solutions to all 12 chapters problem sets are presented in this manual. The proble
gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy
m sets depict examples of practical applications of the concepts described in the book, more
gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy
detailed analysis of some of the ideas, or in some cases present a new concept.
gy gy gy gy gy gy gy gy gy gy gy gy gy gy
Note that selected problems have been given answers already in the book.
gy gy gy gy gy gy gy gy gy gy gy
,1 Chapter One gy
1. Using spherical coordinates, find the capacitance formed by two concentric spherical
gy gy gy gy gy gy gy gy gy gy
conducting shells of radius a, and b. What is the capacitance of a metallic marble wit
gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy
h a diameter of 1cm in free space? Hint: let 𝑏 → ∞, thus, 𝐶 = 4𝜋𝜀𝜀0𝑎 = 0.55𝑝𝐹.
gy gy gy gy gy gy gy gy gy gy gy gy gy gy g y gy gy gy
Solution: Suppose the inner sphere has a surface charge density of +𝜌𝑆. The outer surfa
gy gy gy gy gy gy gy gy gy gy gy gy gy gy
ce charge density is negative, and proportionally smaller (by (𝑎/𝑏)2) to keep the total ch
gy gy gy gy gy gy gy gy gy gy gy gy gy gy
arge the same. gy gy
-
+
+S - + a + -
b
+
-
From Gauss’s law: gy gy
ф𝐷 ⋅ 𝑑𝑆 = 𝑄𝑄 = +𝜌𝑆4𝜋𝑎2
gy gy gy gy gy gy
𝑆
Thus, inside the sphere (𝑎 ≤ 𝑟 ≤
gy gy gy gy gy gy gy gy
𝑏): 𝑎2
𝐷 = 𝜌𝑆 2 𝑎𝑟 gy gy
𝑟 gy
gy
Assuming a potential of 𝑉0 between the inner
𝑎 1 2 and 2 1
outer surfaces, 1we have:
𝜌 𝑎 𝑑𝑟 = 𝜌𝑆 𝑎
gy gy gy gy gy g ygy gy g y gy gy gy gy gy
𝑉 =− gy g y
gy
g y gy gy g y gy g y
0 𝑆 ( − ) gy gyg y g y g
y
2
𝑏 𝑟 𝜖 𝑎 𝑏 gy g
Thus: 𝜖 y
𝜌𝑆4𝜋𝑎2 = 4𝜋𝜖
𝑄𝑄 g y g y
𝐶 =𝑉 =
𝜌 𝑆 21 1 1 1 gy gy gy
g y
g y
𝜖 𝑎 (𝑎 − 𝑏) 𝑎 − 𝑏
0 gy gy
gy gy
1
In the case of a metallic marble, 𝑏 → ∞, and hence: 𝐶 = 4 𝑎. Letting 𝜀𝜀 =
gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy
g y g y gygy
𝜋𝜀𝜀0 0
×
36𝜋
5
10−9, and 𝑎 = 0.5𝑐𝑚, it yields 𝐶 = 𝑝𝐹 = 0.55𝑝𝐹.
gy
gy gy
gy gy gy gy gy gy g y gy g y
9
2. Consider the parallel plate capacitor containing two different dielectrics. Find the tot
gy gy gy gy gy gy gy gy gy gy gy
al capacitance as a function of the parameters shown in the figure.
gy gy gy gy gy gy gy gy gy gy gy
, Area: A gy
1
d1
2
d2
Solution: Since in the boundary no charge exists (perfect insulator), the normal component
gy gy gy gy gy gy gy gy gy gy gy gy
of the electric flux density has to be equal in each dielectric. That is:
gy gy gy gy gy gy gy gy gy gy gy gy gy gy
𝐷1 = 𝐷𝟐𝟐
g y gy
Accordingly:
𝜖1𝐸1 = 𝜖2𝐸𝟐𝟐 g y gy
Assuming a surface charge density of +𝜌𝑆 for the top plate, and −𝜌𝑆 for the bottom plate,
gy gy gy gy gy gy g y gy gy gy gy gy g y gy gy gy
the electric field (or flux has a component only in z direction, and we have:
gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy
𝐷1 = 𝐷𝟐𝟐 = −𝜌𝑆𝑎𝑧
gy gy gy gy
If the potential between the top ad bottom plates is 𝑉0, based on the line integral we obtain:
gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy
𝑑1+𝑑2 𝑑2g y
−𝜌𝑆 𝑑1+𝑑2g y −𝜌 𝜌𝑆 𝜌𝑆
𝑉0 = − 𝐸. 𝑑𝑧 = − 𝑆 𝑑𝑧 = 𝑑1 + 𝑑2
𝜖 gy g y
gy gy gy gy gy gy gy
𝜖 𝜖 gy gy
𝜖
𝑑𝑧 gy
g y g y
−
0 0 2 𝑑2 1 1 2
Since the total charge on each plate is: 𝑄𝑄 = 𝜌𝑆𝐴, the capacitance is found to be:
gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy
= 𝐴 g y
gy
𝐶 = 𝑄𝑄 𝑑 𝑑 gy g y g y
𝑉
0 1g y 2
+ 𝜖2 gy
𝜖1
which is analogous to two parallel capacitor
gy gy gy gy gy gy
s.
3. What would be the capacitance of the structure in problem 2 if there were a third conduc
gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy
tor with zero thickness at the interface of the dielectrics? How would the electric field li
gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy
nes look? How does the capacitance change if the spacing between the top and bottom pla
gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy
tes are kept the same, but the conductor thickness is not zero?
gy gy gy gy gy gy gy gy gy gy gy
The selected solutions to all 12 chapters problem sets are presented in this manual. The proble
gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy
m sets depict examples of practical applications of the concepts described in the book, more
gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy
detailed analysis of some of the ideas, or in some cases present a new concept.
gy gy gy gy gy gy gy gy gy gy gy gy gy gy
Note that selected problems have been given answers already in the book.
gy gy gy gy gy gy gy gy gy gy gy
,1 Chapter One gy
1. Using spherical coordinates, find the capacitance formed by two concentric spherical
gy gy gy gy gy gy gy gy gy gy
conducting shells of radius a, and b. What is the capacitance of a metallic marble wit
gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy
h a diameter of 1cm in free space? Hint: let 𝑏 → ∞, thus, 𝐶 = 4𝜋𝜀𝜀0𝑎 = 0.55𝑝𝐹.
gy gy gy gy gy gy gy gy gy gy gy gy gy gy g y gy gy gy
Solution: Suppose the inner sphere has a surface charge density of +𝜌𝑆. The outer surfa
gy gy gy gy gy gy gy gy gy gy gy gy gy gy
ce charge density is negative, and proportionally smaller (by (𝑎/𝑏)2) to keep the total ch
gy gy gy gy gy gy gy gy gy gy gy gy gy gy
arge the same. gy gy
-
+
+S - + a + -
b
+
-
From Gauss’s law: gy gy
ф𝐷 ⋅ 𝑑𝑆 = 𝑄𝑄 = +𝜌𝑆4𝜋𝑎2
gy gy gy gy gy gy
𝑆
Thus, inside the sphere (𝑎 ≤ 𝑟 ≤
gy gy gy gy gy gy gy gy
𝑏): 𝑎2
𝐷 = 𝜌𝑆 2 𝑎𝑟 gy gy
𝑟 gy
gy
Assuming a potential of 𝑉0 between the inner
𝑎 1 2 and 2 1
outer surfaces, 1we have:
𝜌 𝑎 𝑑𝑟 = 𝜌𝑆 𝑎
gy gy gy gy gy g ygy gy g y gy gy gy gy gy
𝑉 =− gy g y
gy
g y gy gy g y gy g y
0 𝑆 ( − ) gy gyg y g y g
y
2
𝑏 𝑟 𝜖 𝑎 𝑏 gy g
Thus: 𝜖 y
𝜌𝑆4𝜋𝑎2 = 4𝜋𝜖
𝑄𝑄 g y g y
𝐶 =𝑉 =
𝜌 𝑆 21 1 1 1 gy gy gy
g y
g y
𝜖 𝑎 (𝑎 − 𝑏) 𝑎 − 𝑏
0 gy gy
gy gy
1
In the case of a metallic marble, 𝑏 → ∞, and hence: 𝐶 = 4 𝑎. Letting 𝜀𝜀 =
gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy
g y g y gygy
𝜋𝜀𝜀0 0
×
36𝜋
5
10−9, and 𝑎 = 0.5𝑐𝑚, it yields 𝐶 = 𝑝𝐹 = 0.55𝑝𝐹.
gy
gy gy
gy gy gy gy gy gy g y gy g y
9
2. Consider the parallel plate capacitor containing two different dielectrics. Find the tot
gy gy gy gy gy gy gy gy gy gy gy
al capacitance as a function of the parameters shown in the figure.
gy gy gy gy gy gy gy gy gy gy gy
, Area: A gy
1
d1
2
d2
Solution: Since in the boundary no charge exists (perfect insulator), the normal component
gy gy gy gy gy gy gy gy gy gy gy gy
of the electric flux density has to be equal in each dielectric. That is:
gy gy gy gy gy gy gy gy gy gy gy gy gy gy
𝐷1 = 𝐷𝟐𝟐
g y gy
Accordingly:
𝜖1𝐸1 = 𝜖2𝐸𝟐𝟐 g y gy
Assuming a surface charge density of +𝜌𝑆 for the top plate, and −𝜌𝑆 for the bottom plate,
gy gy gy gy gy gy g y gy gy gy gy gy g y gy gy gy
the electric field (or flux has a component only in z direction, and we have:
gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy
𝐷1 = 𝐷𝟐𝟐 = −𝜌𝑆𝑎𝑧
gy gy gy gy
If the potential between the top ad bottom plates is 𝑉0, based on the line integral we obtain:
gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy
𝑑1+𝑑2 𝑑2g y
−𝜌𝑆 𝑑1+𝑑2g y −𝜌 𝜌𝑆 𝜌𝑆
𝑉0 = − 𝐸. 𝑑𝑧 = − 𝑆 𝑑𝑧 = 𝑑1 + 𝑑2
𝜖 gy g y
gy gy gy gy gy gy gy
𝜖 𝜖 gy gy
𝜖
𝑑𝑧 gy
g y g y
−
0 0 2 𝑑2 1 1 2
Since the total charge on each plate is: 𝑄𝑄 = 𝜌𝑆𝐴, the capacitance is found to be:
gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy
= 𝐴 g y
gy
𝐶 = 𝑄𝑄 𝑑 𝑑 gy g y g y
𝑉
0 1g y 2
+ 𝜖2 gy
𝜖1
which is analogous to two parallel capacitor
gy gy gy gy gy gy
s.
3. What would be the capacitance of the structure in problem 2 if there were a third conduc
gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy
tor with zero thickness at the interface of the dielectrics? How would the electric field li
gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy
nes look? How does the capacitance change if the spacing between the top and bottom pla
gy gy gy gy gy gy gy gy gy gy gy gy gy gy gy
tes are kept the same, but the conductor thickness is not zero?
gy gy gy gy gy gy gy gy gy gy gy
