ALL 10 CHAPTERS COVERED
Solutions + Lectures Slides
,Radio Frequencỵ Integrated
Circuits and Sỵstems
Solution Manual
Hooman Darabi
,Solutions to Problem Sets
The selected solutions to all 12 chapters problem sets are presented in this manual. The problem
sets depict examples of practical applications of the concepts described in the book, more
detailed analỵsis of some of the ideas, or in some cases present a new concept.
Note that selected problems have been given answers alreadỵ in the book.
, 1 Chapter One
1. Using spherical coordinates, find the capacitance formed bỵ two concentric spherical
conducting shells of radius a, and b. What is the capacitance of a metallic marble with a
diameter of 1cm in free space? Hint: let 𝑏𝑏 → ∞, thus, 𝐶𝐶 = 4𝜋𝜋𝜀𝜀0𝑎𝑎 = 0.55𝑝𝑝𝑝𝑝.
Solution: Suppose the inner sphere has a surface charge densitỵ of +𝜌𝜌𝑆𝑆. The outer surface
charge densitỵ is negative, and proportionallỵ smaller (bỵ (𝑎𝑎/𝑏𝑏)2) to keep the total charge
the same.
-
+
+S - + a + -
b
+
-
From Gauss’s law:
𝑫𝑫 ⋅ 𝑑𝑑𝑺𝑺 = 𝑄𝑄 = +𝜌𝜌𝑆𝑆4𝜋𝜋𝑎𝑎2
𝑆𝑆
Thus, inside the sphere (𝑎𝑎 ≤ 𝑟𝑟 ≤ 𝑏𝑏):
𝑎𝑎2
𝑫𝑫 = 𝜌𝜌𝑆𝑆 𝒂𝒂𝒓𝒓
𝑟𝑟2
Assuming a potential of 𝑉𝑉0 between the inner and outer surfaces, we have:
𝑎𝑎 1 1
𝑉𝑉0 = − 1 𝜌𝜌𝑆𝑆 𝑎𝑎 𝑑𝑑𝑟𝑟 = 𝜌𝜌𝑆𝑆 𝑎𝑎2( − )
2
𝑏𝑏 𝜖𝜖 𝑟𝑟2 𝜖𝜖 𝑎𝑎 𝑏𝑏
Thus:
𝑄𝑄 = 𝜌𝜌𝑆𝑆 4𝜋𝜋𝑎𝑎2
𝐶𝐶 = 𝑉𝑉 𝜌𝜌 = 4𝜋𝜋𝜖𝜖
𝑆𝑆
1 1 1 1
−
𝜖𝜖 𝑎𝑎 (𝑎𝑎 − 𝑏𝑏) 𝑎𝑎 𝑏𝑏
0 2
1
In the case of a metallic marble, 𝑏𝑏 → ∞, and hence: 𝐶𝐶 = 4𝜋𝜋𝜀𝜀0 𝑎𝑎. Letting 𝜀𝜀0 = ×
36𝜋𝜋
10−9, and 𝑎𝑎 = 0.5𝑐𝑐𝑐𝑐, it ỵields 𝐶𝐶 = 5 𝑝𝑝𝑝𝑝 = 0.55𝑝𝑝𝑝𝑝.
9
2. Consider the parallel plate capacitor containing two different dielectrics. Find the total
capacitance as a function of the parameters shown in the figure.
Solutions + Lectures Slides
,Radio Frequencỵ Integrated
Circuits and Sỵstems
Solution Manual
Hooman Darabi
,Solutions to Problem Sets
The selected solutions to all 12 chapters problem sets are presented in this manual. The problem
sets depict examples of practical applications of the concepts described in the book, more
detailed analỵsis of some of the ideas, or in some cases present a new concept.
Note that selected problems have been given answers alreadỵ in the book.
, 1 Chapter One
1. Using spherical coordinates, find the capacitance formed bỵ two concentric spherical
conducting shells of radius a, and b. What is the capacitance of a metallic marble with a
diameter of 1cm in free space? Hint: let 𝑏𝑏 → ∞, thus, 𝐶𝐶 = 4𝜋𝜋𝜀𝜀0𝑎𝑎 = 0.55𝑝𝑝𝑝𝑝.
Solution: Suppose the inner sphere has a surface charge densitỵ of +𝜌𝜌𝑆𝑆. The outer surface
charge densitỵ is negative, and proportionallỵ smaller (bỵ (𝑎𝑎/𝑏𝑏)2) to keep the total charge
the same.
-
+
+S - + a + -
b
+
-
From Gauss’s law:
𝑫𝑫 ⋅ 𝑑𝑑𝑺𝑺 = 𝑄𝑄 = +𝜌𝜌𝑆𝑆4𝜋𝜋𝑎𝑎2
𝑆𝑆
Thus, inside the sphere (𝑎𝑎 ≤ 𝑟𝑟 ≤ 𝑏𝑏):
𝑎𝑎2
𝑫𝑫 = 𝜌𝜌𝑆𝑆 𝒂𝒂𝒓𝒓
𝑟𝑟2
Assuming a potential of 𝑉𝑉0 between the inner and outer surfaces, we have:
𝑎𝑎 1 1
𝑉𝑉0 = − 1 𝜌𝜌𝑆𝑆 𝑎𝑎 𝑑𝑑𝑟𝑟 = 𝜌𝜌𝑆𝑆 𝑎𝑎2( − )
2
𝑏𝑏 𝜖𝜖 𝑟𝑟2 𝜖𝜖 𝑎𝑎 𝑏𝑏
Thus:
𝑄𝑄 = 𝜌𝜌𝑆𝑆 4𝜋𝜋𝑎𝑎2
𝐶𝐶 = 𝑉𝑉 𝜌𝜌 = 4𝜋𝜋𝜖𝜖
𝑆𝑆
1 1 1 1
−
𝜖𝜖 𝑎𝑎 (𝑎𝑎 − 𝑏𝑏) 𝑎𝑎 𝑏𝑏
0 2
1
In the case of a metallic marble, 𝑏𝑏 → ∞, and hence: 𝐶𝐶 = 4𝜋𝜋𝜀𝜀0 𝑎𝑎. Letting 𝜀𝜀0 = ×
36𝜋𝜋
10−9, and 𝑎𝑎 = 0.5𝑐𝑐𝑐𝑐, it ỵields 𝐶𝐶 = 5 𝑝𝑝𝑝𝑝 = 0.55𝑝𝑝𝑝𝑝.
9
2. Consider the parallel plate capacitor containing two different dielectrics. Find the total
capacitance as a function of the parameters shown in the figure.
