,Radio Frequency Integrated
Circuits and Systems
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 analysis of some of the ideas, or in some cases present a new concept.
Note that selected problems have been given answers already in the book.
,1 Chapter One
1. Using spherical coordinates, find the capacitance formed by 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 = 4𝜋𝜋𝜀𝜀0 𝑎𝑎 = 0.55𝑝𝑝𝑝𝑝.
Solution: Suppose the inner sphere has a surface charge density of +𝜌𝜌𝑆𝑆 . The outer surface
charge density is negative, and proportionally smaller (by (𝑎𝑎/𝑏𝑏)2) to keep the total charge
the same.
-
+
+ρS - + a + -
b
+
-
From Gauss’s law:
�𝑫𝑫 ⋅ 𝑑𝑑𝑺𝑺 = 𝑄𝑄 = +𝜌𝜌𝑆𝑆 4𝜋𝜋𝑎𝑎2
𝑆𝑆
Thus, inside the sphere (𝑎𝑎 ≤ 𝑟𝑟 ≤ 𝑏𝑏):
𝑎𝑎2
𝑫𝑫 = 𝜌𝜌𝑆𝑆
𝒂𝒂
𝑟𝑟 2 𝒓𝒓
Assuming a potential of 𝑉𝑉0 between the inner and outer surfaces, we have:
𝑎𝑎
1 𝑎𝑎2 𝜌𝜌𝑆𝑆 1 1
𝑉𝑉0 = − � 𝜌𝜌𝑆𝑆 2 𝑑𝑑𝑑𝑑 = 𝑎𝑎2 ( − )
𝑏𝑏 𝜖𝜖 𝑟𝑟 𝜖𝜖 𝑎𝑎 𝑏𝑏
Thus:
𝑄𝑄 𝜌𝜌𝑆𝑆 4𝜋𝜋𝑎𝑎2 4𝜋𝜋𝜋𝜋
𝐶𝐶 = = =
𝑉𝑉0 𝜌𝜌𝑆𝑆 𝑎𝑎2 (1 − 1) 1 − 1
𝜖𝜖 𝑎𝑎 𝑏𝑏 𝑎𝑎 𝑏𝑏
1
In the case of a metallic marble, 𝑏𝑏 → ∞, and hence: 𝐶𝐶 = 4𝜋𝜋𝜀𝜀0 𝑎𝑎. Letting 𝜀𝜀0 = 36𝜋𝜋 ×
5
10−9 , and 𝑎𝑎 = 0.5𝑐𝑐𝑐𝑐, it yields 𝐶𝐶 = 9 𝑝𝑝𝑝𝑝 = 0.55𝑝𝑝𝑝𝑝.
2. Consider the parallel plate capacitor containing two different dielectrics. Find the total
capacitance as a function of the parameters shown in the figure.
, Area: A
ε1
d1
ε2
d2
Solution: Since in the boundary no charge exists (perfect insulator), the normal component
of the electric flux density has to be equal in each dielectric. That is:
𝑫𝑫𝟏𝟏 = 𝑫𝑫𝟐𝟐
Accordingly:
𝜖𝜖1 𝑬𝑬𝟏𝟏 = 𝜖𝜖2 𝑬𝑬𝟐𝟐
Assuming a surface charge density of +𝜌𝜌𝑆𝑆 for the top plate, and −𝜌𝜌𝑆𝑆 for the bottom plate, the
electric field (or flux has a component only in z direction, and we have:
𝑫𝑫𝟏𝟏 = 𝑫𝑫𝟐𝟐 = −𝜌𝜌𝑆𝑆 𝒂𝒂𝒛𝒛
If the potential between the top ad bottom plates is 𝑉𝑉0, based on the line integral we obtain:
𝑑𝑑1 +𝑑𝑑2 𝑑𝑑2 𝑑𝑑1 +𝑑𝑑2
−𝜌𝜌𝑆𝑆 −𝜌𝜌𝑆𝑆 𝜌𝜌𝑆𝑆 𝜌𝜌𝑆𝑆
𝑉𝑉0 = − � 𝑬𝑬. 𝑑𝑑𝒛𝒛 = − � 𝑑𝑑𝑑𝑑 − � 𝑑𝑑𝑑𝑑 = 𝑑𝑑1 + 𝑑𝑑2
0 0 𝜖𝜖2 𝑑𝑑2 𝜖𝜖1 𝜖𝜖1 𝜖𝜖2
Since the total charge on each plate is: 𝑄𝑄 = 𝜌𝜌𝑆𝑆 𝐴𝐴, the capacitance is found to be:
𝑄𝑄 𝐴𝐴
𝐶𝐶 = =
𝑉𝑉0 𝑑𝑑1 + 𝑑𝑑2
𝜖𝜖1 𝜖𝜖2
which is analogous to two parallel capacitors.
3. What would be the capacitance of the structure in problem 2 if there were a third conductor
with zero thickness at the interface of the dielectrics? How would the electric field lines
look? How does the capacitance change if the spacing between the top and bottom plates are
kept the same, but the conductor thickness is not zero?
, Solution: If the conductor is perfect, opposite charges are formed on the surface, but the
capacitance remains the same, that is to say, the electric fields terminate to the conductor, but
are not altered.
If the conductor thickness is greater than zero, but the total distance between the top and
bottom plates is the same (𝑑𝑑1 + 𝑑𝑑2 ), we expect the capacitance to increase.
4. Repeat problem 2 if the dielectric boundary were placed normal to the two conducting plates
as shown below.
d
A1 A2
ε1 ε2
Solution: Similar to 2, the electric flux density is in z direction, and we assume a surface
charge density of +𝜌𝜌𝑆𝑆1/2 for the top plates, and −𝜌𝜌𝑆𝑆1/2 for the bottom plates. Assuming a
potential of 𝑉𝑉0 between the plates, unlike 2, as 𝑫𝑫 is tangent to the surface, in general 𝑫𝑫𝟏𝟏 ≠
𝑫𝑫𝟐𝟐 . Thus, we do not assume a uniform charge density on the plates. Furthermore, based on
the line integral definition, at the boundary the tangent components of the electric field
(which are in z direction) must be equal between the two dielectrics, that is:
𝑬𝑬𝟏𝟏 = 𝑬𝑬𝟐𝟐
which yields:
𝜌𝜌𝑆𝑆1 𝜌𝜌𝑆𝑆2
=
𝜖𝜖1 𝜖𝜖2
Finally, for the potential the line integral yields:
𝜌𝜌𝑆𝑆1 𝜌𝜌𝑆𝑆2
𝑉𝑉0 = 𝑑𝑑 = 𝑑𝑑
𝜖𝜖1 𝜖𝜖2
The total charge is: 𝑄𝑄 = 𝜌𝜌𝑆𝑆1 𝐴𝐴1 + 𝜌𝜌𝑆𝑆2 𝐴𝐴2
Consequently:
𝑄𝑄 𝜖𝜖1 𝐴𝐴1 + 𝜖𝜖2 𝐴𝐴2
𝐶𝐶 = =
𝑉𝑉0 𝑑𝑑
, As expected, this case turns out to be similar to two series capacitances.
5. Analogues to the capacitance, using Ohm’s law, show that the leakage conductance of an
∫ 𝐄𝐄⋅𝑑𝑑𝐒𝐒
almost perfect conductor with a non-infinite conductivity of σ is given by: 𝐺𝐺 = 𝜎𝜎 −𝑆𝑆 .
∫ 𝑬𝑬.𝑑𝑑𝑳𝑳
Calculate the leakage conductance of a coaxial cable with radii a and b as was used
throughout the chapter.
Solution: In a given conductor we have:
𝐼𝐼 = �𝐉𝐉 ⋅ 𝑑𝑑𝐒𝐒
𝑆𝑆
where 𝐉𝐉 is the current density, and by definition, for a conductor: 𝐉𝐉 = σ𝐄𝐄. According to
Ohm’s law:
𝐼𝐼 ∫ 𝐉𝐉 ⋅ 𝑑𝑑𝐒𝐒 ∫ 𝐄𝐄 ⋅ 𝑑𝑑𝐒𝐒
𝐺𝐺 = = 𝑆𝑆 = 𝜎𝜎 𝑆𝑆
𝑉𝑉 − ∫ 𝑬𝑬. 𝑑𝑑𝑳𝑳 − ∫ 𝑬𝑬. 𝑑𝑑𝑳𝑳
which has a similar form as the capacitance equation:
𝑄𝑄 ∮ 𝑬𝑬 ⋅ 𝑑𝑑𝑺𝑺
𝐶𝐶 = = 𝜖𝜖 𝑆𝑆
𝑉𝑉 − ∫ 𝑬𝑬. 𝑑𝑑𝑳𝑳
Note that the surface integral in the capacitance equation is over a closed surface.
6. Consider a very long hollow charge-free super conductor cylindrical shell with inner and outer
radios of a and b, respectively. A wire with a current I is placed at the center of the cylinder.
Calculate the magnetic field inside and outside considering that the magnetic field inside the
shell would have to be zero. If the current I is moved away from the center but inside the shell,
how the magnetic fields inside and outside would alter?
I
a
b
, Solution: Based on Ampere’s law, for 𝑟𝑟 ≤ 𝑎𝑎 we have:
� 𝑯𝑯 ⋅ 𝑑𝑑𝑳𝑳 = 𝐼𝐼
Therefore:
𝐼𝐼
𝑯𝑯 =
𝒂𝒂
2𝜋𝜋𝜋𝜋 𝝓𝝓
For (𝑎𝑎 ≤ 𝑟𝑟 ≤ 𝑏𝑏), that is inside the superconductor, the magnetic field (and flux) are zero. In
practice, the magnetic flux needs to be constant, so that the voltage is zero. Otherwise, there
will be an infinite current induced in the superconductor. In practice however, any small
change in magnetic flux will induce an infinite current, and thus, 𝑩𝑩 = 0. Furthermore, a
−𝐼𝐼
surface current of 2𝜋𝜋𝜋𝜋 𝒂𝒂𝒛𝒛 flows on the inner surface,
𝐼𝐼
Outside the conductor (𝑟𝑟 ≥ 𝑏𝑏), a surface current of 2𝜋𝜋𝜋𝜋 𝒂𝒂𝒛𝒛 flows on the outer surface, and
𝐼𝐼
again, 𝑯𝑯 = 2𝜋𝜋𝜋𝜋 𝒂𝒂𝝓𝝓 .
If the current moves away from the center, 𝑩𝑩 changes for 𝑟𝑟 ≤ 𝑎𝑎, but remains the same
outside the conductor. The surface current on the inner shell is not uniform anymore, but
remains the same for the outer shell.
7. What is the internal inductance (per length) of a long straight wire with a circular cross
𝜇𝜇
section of radius a (use energy definition)? Answer: 8𝜋𝜋0 .
Solution: Due to symmetry, we can argue that the magnetic field has only a component in
the 𝒂𝒂𝝓𝝓 direction. The current density inside the wire (𝑟𝑟 ≤ 𝑎𝑎) is:
𝜋𝜋𝑟𝑟 2 𝑟𝑟 2
𝑲𝑲 = 𝐼𝐼 2 𝒂𝒂𝒛𝒛 = 𝐼𝐼 2 𝒂𝒂𝒛𝒛
𝜋𝜋𝑎𝑎 𝑎𝑎
Accordingly, based on Ampere’s law, the magnetic field is found to be:
𝑟𝑟 2
𝐼𝐼 2 𝑟𝑟
𝑯𝑯 = 𝑎𝑎 𝒂𝒂𝝓𝝓 = 𝐼𝐼 𝒂𝒂
2𝜋𝜋𝜋𝜋 2𝜋𝜋𝑎𝑎2 𝝓𝝓
Next, we shall find the magnetic energy per unit length inside the wire:
𝜇𝜇0 𝟐𝟐
𝜇𝜇0 1 𝑎𝑎 2𝜋𝜋 𝑟𝑟 2 𝜇𝜇0 2
𝑊𝑊𝐻𝐻 = � |𝐇𝐇| dV = � � � (𝐼𝐼 ) 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 = 𝐼𝐼
2 𝑉𝑉 2 0 0 0 2𝜋𝜋𝑎𝑎2 16𝜋𝜋
1
Equating the energy to: 2 𝐿𝐿𝐼𝐼 2 , we obtain the inductance per unit length:
𝜇𝜇0
𝐿𝐿 =
8𝜋𝜋
8. Show that the DC inductance of a piece of wire with finite length l and radius r is: 𝐿𝐿 =
𝜇𝜇0 𝑙𝑙 2𝑙𝑙 3
2𝜋𝜋
(𝑙𝑙𝑙𝑙 𝑟𝑟
− 4). What is the inductance of a copper bond-wire with length of 2mm and a
diameter of 25µm (practical bonding pads in integrated circuits are typically 50×50µm2)?
, Argue why traditionally, as a rule of thumb an inductance of 1nH/mm is assumed for bond-
wires.
Solution: The inductance calculation is detailed by Rosa 1. There are two parts, the internal
𝜇𝜇
inductance, 𝐿𝐿𝑖𝑖𝑖𝑖𝑖𝑖 , which was calculated to be 𝐿𝐿𝑖𝑖𝑖𝑖𝑖𝑖 = 8𝜋𝜋0 𝑙𝑙 in the previous problem, and the
external inductance.
As for the external inductance, let us first find the magnetic field. From the law of Biot-
Savart, the magnetic field at a point P normal to the paper due to an element of length 𝑑𝑑𝑑𝑑 is:
𝐼𝐼𝐼𝐼𝐼𝐼 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼
𝑑𝑑𝑑𝑑 = 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =
4𝜋𝜋𝑅𝑅 2 4𝜋𝜋(𝑥𝑥 2 + (𝑦𝑦 − 𝑏𝑏)2 )3/2
where 𝐼𝐼 is the wire current uniformly distributed, and the rest of the parameters are shown in
the figure below.
Wire
dx
dy
R
θ l
P
y
b
x
The magnetic field at P due to the entire length of the wire is then:
𝑙𝑙
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝐼𝐼 𝑙𝑙 − 𝑏𝑏 𝑏𝑏
𝐻𝐻 = � 2 2 3/2
= ( + )
0 4𝜋𝜋(𝑥𝑥 + (𝑦𝑦 − 𝑏𝑏) ) 4𝜋𝜋𝜋𝜋 �𝑥𝑥 2 + (𝑙𝑙 − 𝑏𝑏)2 √𝑥𝑥 2 + 𝑏𝑏 2
𝐼𝐼
If the integral were to be taken from −∞ to +∞, the field would be 2𝜋𝜋𝜋𝜋 as we calculated
before for a piece of wire with infinite length. To find the inductance, we calculate the
magnetic flux as follows:
𝜇𝜇0 𝐼𝐼 ∞ 𝑙𝑙 𝑙𝑙 − 𝑏𝑏 𝑏𝑏
𝜙𝜙 = �𝐁𝐁 ⋅ 𝑑𝑑𝐒𝐒 = � � ( + ) 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
𝑆𝑆 4𝜋𝜋 𝑥𝑥=𝑟𝑟 𝑏𝑏=0 𝑥𝑥�𝑥𝑥 2 + (𝑙𝑙 − 𝑏𝑏)2 𝑥𝑥√𝑥𝑥 2 + 𝑏𝑏 2
1 Edward B. Rosa, Bulletin of the Bureau of Standards, vol. 4, no.2 , pp 301-305, 1907.
, which is found to be:
𝜇𝜇0 𝐼𝐼 𝑙𝑙 + √𝑟𝑟 2 + 𝑙𝑙 2 𝑟𝑟 √𝑟𝑟 2 + 𝑙𝑙 2
𝜙𝜙 = 𝑙𝑙[𝑙𝑙𝑙𝑙 + − ]
2𝜋𝜋 𝑟𝑟 𝑙𝑙 𝑙𝑙
From this the external inductance is:
𝜇𝜇0 𝑙𝑙 + √𝑟𝑟 2 + 𝑙𝑙 2 𝑟𝑟 √𝑟𝑟 2 + 𝑙𝑙 2
𝐿𝐿𝑒𝑒𝑒𝑒𝑒𝑒 =
𝑙𝑙[𝑙𝑙𝑙𝑙 + − ]
2𝜋𝜋 𝑟𝑟 𝑙𝑙 𝑙𝑙
And the total inductance would be:
𝜇𝜇0 𝑙𝑙 + √𝑟𝑟 2 + 𝑙𝑙 2 𝑟𝑟 1 √𝑟𝑟 2 + 𝑙𝑙 2
𝐿𝐿 = 𝐿𝐿𝑖𝑖𝑖𝑖𝑖𝑖 + 𝐿𝐿𝑒𝑒𝑒𝑒𝑒𝑒 =
𝑙𝑙[𝑙𝑙𝑙𝑙 + + − ]
2𝜋𝜋 𝑟𝑟 𝑙𝑙 4 𝑙𝑙
For 𝑟𝑟 ≪ 𝑙𝑙, the inductance is roughly:
𝜇𝜇0 2𝑙𝑙 3
𝐿𝐿 ≈ 𝑙𝑙(𝑙𝑙𝑙𝑙 − )
2𝜋𝜋 𝑟𝑟 4
For typical values of 𝑟𝑟 = 12.5𝜇𝜇𝜇𝜇 , and 𝑙𝑙 = 2𝑚𝑚𝑚𝑚, the inductance is found to be about
2𝑙𝑙
2.01nH. Given the logarithmic nature of the term 𝑙𝑙𝑙𝑙 𝑟𝑟 , as a rule of thumb we assign an
inductance of about 1nH/mm for a piece of wire. For the reference, a 1mm long wire
inductance is 0.87nH.
9. In Faraday’s experiment, assume the switch has a resistance of R, and the two coils are
identical with an inductance of L. The battery voltage is VBAT. Find the time-varying current
in the coil. Assuming the iron toroid has a large permeability, find the magnetic flux in the
second coil and estimate the emf read by the galvanometer.
Solution: We use the following circuit model to obtain the current in the primary:
R
VBAT
emf
VBAT i1 L i2
t
Circuits and Systems
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 analysis of some of the ideas, or in some cases present a new concept.
Note that selected problems have been given answers already in the book.
,1 Chapter One
1. Using spherical coordinates, find the capacitance formed by 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 = 4𝜋𝜋𝜀𝜀0 𝑎𝑎 = 0.55𝑝𝑝𝑝𝑝.
Solution: Suppose the inner sphere has a surface charge density of +𝜌𝜌𝑆𝑆 . The outer surface
charge density is negative, and proportionally smaller (by (𝑎𝑎/𝑏𝑏)2) to keep the total charge
the same.
-
+
+ρS - + a + -
b
+
-
From Gauss’s law:
�𝑫𝑫 ⋅ 𝑑𝑑𝑺𝑺 = 𝑄𝑄 = +𝜌𝜌𝑆𝑆 4𝜋𝜋𝑎𝑎2
𝑆𝑆
Thus, inside the sphere (𝑎𝑎 ≤ 𝑟𝑟 ≤ 𝑏𝑏):
𝑎𝑎2
𝑫𝑫 = 𝜌𝜌𝑆𝑆
𝒂𝒂
𝑟𝑟 2 𝒓𝒓
Assuming a potential of 𝑉𝑉0 between the inner and outer surfaces, we have:
𝑎𝑎
1 𝑎𝑎2 𝜌𝜌𝑆𝑆 1 1
𝑉𝑉0 = − � 𝜌𝜌𝑆𝑆 2 𝑑𝑑𝑑𝑑 = 𝑎𝑎2 ( − )
𝑏𝑏 𝜖𝜖 𝑟𝑟 𝜖𝜖 𝑎𝑎 𝑏𝑏
Thus:
𝑄𝑄 𝜌𝜌𝑆𝑆 4𝜋𝜋𝑎𝑎2 4𝜋𝜋𝜋𝜋
𝐶𝐶 = = =
𝑉𝑉0 𝜌𝜌𝑆𝑆 𝑎𝑎2 (1 − 1) 1 − 1
𝜖𝜖 𝑎𝑎 𝑏𝑏 𝑎𝑎 𝑏𝑏
1
In the case of a metallic marble, 𝑏𝑏 → ∞, and hence: 𝐶𝐶 = 4𝜋𝜋𝜀𝜀0 𝑎𝑎. Letting 𝜀𝜀0 = 36𝜋𝜋 ×
5
10−9 , and 𝑎𝑎 = 0.5𝑐𝑐𝑐𝑐, it yields 𝐶𝐶 = 9 𝑝𝑝𝑝𝑝 = 0.55𝑝𝑝𝑝𝑝.
2. Consider the parallel plate capacitor containing two different dielectrics. Find the total
capacitance as a function of the parameters shown in the figure.
, Area: A
ε1
d1
ε2
d2
Solution: Since in the boundary no charge exists (perfect insulator), the normal component
of the electric flux density has to be equal in each dielectric. That is:
𝑫𝑫𝟏𝟏 = 𝑫𝑫𝟐𝟐
Accordingly:
𝜖𝜖1 𝑬𝑬𝟏𝟏 = 𝜖𝜖2 𝑬𝑬𝟐𝟐
Assuming a surface charge density of +𝜌𝜌𝑆𝑆 for the top plate, and −𝜌𝜌𝑆𝑆 for the bottom plate, the
electric field (or flux has a component only in z direction, and we have:
𝑫𝑫𝟏𝟏 = 𝑫𝑫𝟐𝟐 = −𝜌𝜌𝑆𝑆 𝒂𝒂𝒛𝒛
If the potential between the top ad bottom plates is 𝑉𝑉0, based on the line integral we obtain:
𝑑𝑑1 +𝑑𝑑2 𝑑𝑑2 𝑑𝑑1 +𝑑𝑑2
−𝜌𝜌𝑆𝑆 −𝜌𝜌𝑆𝑆 𝜌𝜌𝑆𝑆 𝜌𝜌𝑆𝑆
𝑉𝑉0 = − � 𝑬𝑬. 𝑑𝑑𝒛𝒛 = − � 𝑑𝑑𝑑𝑑 − � 𝑑𝑑𝑑𝑑 = 𝑑𝑑1 + 𝑑𝑑2
0 0 𝜖𝜖2 𝑑𝑑2 𝜖𝜖1 𝜖𝜖1 𝜖𝜖2
Since the total charge on each plate is: 𝑄𝑄 = 𝜌𝜌𝑆𝑆 𝐴𝐴, the capacitance is found to be:
𝑄𝑄 𝐴𝐴
𝐶𝐶 = =
𝑉𝑉0 𝑑𝑑1 + 𝑑𝑑2
𝜖𝜖1 𝜖𝜖2
which is analogous to two parallel capacitors.
3. What would be the capacitance of the structure in problem 2 if there were a third conductor
with zero thickness at the interface of the dielectrics? How would the electric field lines
look? How does the capacitance change if the spacing between the top and bottom plates are
kept the same, but the conductor thickness is not zero?
, Solution: If the conductor is perfect, opposite charges are formed on the surface, but the
capacitance remains the same, that is to say, the electric fields terminate to the conductor, but
are not altered.
If the conductor thickness is greater than zero, but the total distance between the top and
bottom plates is the same (𝑑𝑑1 + 𝑑𝑑2 ), we expect the capacitance to increase.
4. Repeat problem 2 if the dielectric boundary were placed normal to the two conducting plates
as shown below.
d
A1 A2
ε1 ε2
Solution: Similar to 2, the electric flux density is in z direction, and we assume a surface
charge density of +𝜌𝜌𝑆𝑆1/2 for the top plates, and −𝜌𝜌𝑆𝑆1/2 for the bottom plates. Assuming a
potential of 𝑉𝑉0 between the plates, unlike 2, as 𝑫𝑫 is tangent to the surface, in general 𝑫𝑫𝟏𝟏 ≠
𝑫𝑫𝟐𝟐 . Thus, we do not assume a uniform charge density on the plates. Furthermore, based on
the line integral definition, at the boundary the tangent components of the electric field
(which are in z direction) must be equal between the two dielectrics, that is:
𝑬𝑬𝟏𝟏 = 𝑬𝑬𝟐𝟐
which yields:
𝜌𝜌𝑆𝑆1 𝜌𝜌𝑆𝑆2
=
𝜖𝜖1 𝜖𝜖2
Finally, for the potential the line integral yields:
𝜌𝜌𝑆𝑆1 𝜌𝜌𝑆𝑆2
𝑉𝑉0 = 𝑑𝑑 = 𝑑𝑑
𝜖𝜖1 𝜖𝜖2
The total charge is: 𝑄𝑄 = 𝜌𝜌𝑆𝑆1 𝐴𝐴1 + 𝜌𝜌𝑆𝑆2 𝐴𝐴2
Consequently:
𝑄𝑄 𝜖𝜖1 𝐴𝐴1 + 𝜖𝜖2 𝐴𝐴2
𝐶𝐶 = =
𝑉𝑉0 𝑑𝑑
, As expected, this case turns out to be similar to two series capacitances.
5. Analogues to the capacitance, using Ohm’s law, show that the leakage conductance of an
∫ 𝐄𝐄⋅𝑑𝑑𝐒𝐒
almost perfect conductor with a non-infinite conductivity of σ is given by: 𝐺𝐺 = 𝜎𝜎 −𝑆𝑆 .
∫ 𝑬𝑬.𝑑𝑑𝑳𝑳
Calculate the leakage conductance of a coaxial cable with radii a and b as was used
throughout the chapter.
Solution: In a given conductor we have:
𝐼𝐼 = �𝐉𝐉 ⋅ 𝑑𝑑𝐒𝐒
𝑆𝑆
where 𝐉𝐉 is the current density, and by definition, for a conductor: 𝐉𝐉 = σ𝐄𝐄. According to
Ohm’s law:
𝐼𝐼 ∫ 𝐉𝐉 ⋅ 𝑑𝑑𝐒𝐒 ∫ 𝐄𝐄 ⋅ 𝑑𝑑𝐒𝐒
𝐺𝐺 = = 𝑆𝑆 = 𝜎𝜎 𝑆𝑆
𝑉𝑉 − ∫ 𝑬𝑬. 𝑑𝑑𝑳𝑳 − ∫ 𝑬𝑬. 𝑑𝑑𝑳𝑳
which has a similar form as the capacitance equation:
𝑄𝑄 ∮ 𝑬𝑬 ⋅ 𝑑𝑑𝑺𝑺
𝐶𝐶 = = 𝜖𝜖 𝑆𝑆
𝑉𝑉 − ∫ 𝑬𝑬. 𝑑𝑑𝑳𝑳
Note that the surface integral in the capacitance equation is over a closed surface.
6. Consider a very long hollow charge-free super conductor cylindrical shell with inner and outer
radios of a and b, respectively. A wire with a current I is placed at the center of the cylinder.
Calculate the magnetic field inside and outside considering that the magnetic field inside the
shell would have to be zero. If the current I is moved away from the center but inside the shell,
how the magnetic fields inside and outside would alter?
I
a
b
, Solution: Based on Ampere’s law, for 𝑟𝑟 ≤ 𝑎𝑎 we have:
� 𝑯𝑯 ⋅ 𝑑𝑑𝑳𝑳 = 𝐼𝐼
Therefore:
𝐼𝐼
𝑯𝑯 =
𝒂𝒂
2𝜋𝜋𝜋𝜋 𝝓𝝓
For (𝑎𝑎 ≤ 𝑟𝑟 ≤ 𝑏𝑏), that is inside the superconductor, the magnetic field (and flux) are zero. In
practice, the magnetic flux needs to be constant, so that the voltage is zero. Otherwise, there
will be an infinite current induced in the superconductor. In practice however, any small
change in magnetic flux will induce an infinite current, and thus, 𝑩𝑩 = 0. Furthermore, a
−𝐼𝐼
surface current of 2𝜋𝜋𝜋𝜋 𝒂𝒂𝒛𝒛 flows on the inner surface,
𝐼𝐼
Outside the conductor (𝑟𝑟 ≥ 𝑏𝑏), a surface current of 2𝜋𝜋𝜋𝜋 𝒂𝒂𝒛𝒛 flows on the outer surface, and
𝐼𝐼
again, 𝑯𝑯 = 2𝜋𝜋𝜋𝜋 𝒂𝒂𝝓𝝓 .
If the current moves away from the center, 𝑩𝑩 changes for 𝑟𝑟 ≤ 𝑎𝑎, but remains the same
outside the conductor. The surface current on the inner shell is not uniform anymore, but
remains the same for the outer shell.
7. What is the internal inductance (per length) of a long straight wire with a circular cross
𝜇𝜇
section of radius a (use energy definition)? Answer: 8𝜋𝜋0 .
Solution: Due to symmetry, we can argue that the magnetic field has only a component in
the 𝒂𝒂𝝓𝝓 direction. The current density inside the wire (𝑟𝑟 ≤ 𝑎𝑎) is:
𝜋𝜋𝑟𝑟 2 𝑟𝑟 2
𝑲𝑲 = 𝐼𝐼 2 𝒂𝒂𝒛𝒛 = 𝐼𝐼 2 𝒂𝒂𝒛𝒛
𝜋𝜋𝑎𝑎 𝑎𝑎
Accordingly, based on Ampere’s law, the magnetic field is found to be:
𝑟𝑟 2
𝐼𝐼 2 𝑟𝑟
𝑯𝑯 = 𝑎𝑎 𝒂𝒂𝝓𝝓 = 𝐼𝐼 𝒂𝒂
2𝜋𝜋𝜋𝜋 2𝜋𝜋𝑎𝑎2 𝝓𝝓
Next, we shall find the magnetic energy per unit length inside the wire:
𝜇𝜇0 𝟐𝟐
𝜇𝜇0 1 𝑎𝑎 2𝜋𝜋 𝑟𝑟 2 𝜇𝜇0 2
𝑊𝑊𝐻𝐻 = � |𝐇𝐇| dV = � � � (𝐼𝐼 ) 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 = 𝐼𝐼
2 𝑉𝑉 2 0 0 0 2𝜋𝜋𝑎𝑎2 16𝜋𝜋
1
Equating the energy to: 2 𝐿𝐿𝐼𝐼 2 , we obtain the inductance per unit length:
𝜇𝜇0
𝐿𝐿 =
8𝜋𝜋
8. Show that the DC inductance of a piece of wire with finite length l and radius r is: 𝐿𝐿 =
𝜇𝜇0 𝑙𝑙 2𝑙𝑙 3
2𝜋𝜋
(𝑙𝑙𝑙𝑙 𝑟𝑟
− 4). What is the inductance of a copper bond-wire with length of 2mm and a
diameter of 25µm (practical bonding pads in integrated circuits are typically 50×50µm2)?
, Argue why traditionally, as a rule of thumb an inductance of 1nH/mm is assumed for bond-
wires.
Solution: The inductance calculation is detailed by Rosa 1. There are two parts, the internal
𝜇𝜇
inductance, 𝐿𝐿𝑖𝑖𝑖𝑖𝑖𝑖 , which was calculated to be 𝐿𝐿𝑖𝑖𝑖𝑖𝑖𝑖 = 8𝜋𝜋0 𝑙𝑙 in the previous problem, and the
external inductance.
As for the external inductance, let us first find the magnetic field. From the law of Biot-
Savart, the magnetic field at a point P normal to the paper due to an element of length 𝑑𝑑𝑑𝑑 is:
𝐼𝐼𝐼𝐼𝐼𝐼 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼
𝑑𝑑𝑑𝑑 = 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =
4𝜋𝜋𝑅𝑅 2 4𝜋𝜋(𝑥𝑥 2 + (𝑦𝑦 − 𝑏𝑏)2 )3/2
where 𝐼𝐼 is the wire current uniformly distributed, and the rest of the parameters are shown in
the figure below.
Wire
dx
dy
R
θ l
P
y
b
x
The magnetic field at P due to the entire length of the wire is then:
𝑙𝑙
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝐼𝐼 𝑙𝑙 − 𝑏𝑏 𝑏𝑏
𝐻𝐻 = � 2 2 3/2
= ( + )
0 4𝜋𝜋(𝑥𝑥 + (𝑦𝑦 − 𝑏𝑏) ) 4𝜋𝜋𝜋𝜋 �𝑥𝑥 2 + (𝑙𝑙 − 𝑏𝑏)2 √𝑥𝑥 2 + 𝑏𝑏 2
𝐼𝐼
If the integral were to be taken from −∞ to +∞, the field would be 2𝜋𝜋𝜋𝜋 as we calculated
before for a piece of wire with infinite length. To find the inductance, we calculate the
magnetic flux as follows:
𝜇𝜇0 𝐼𝐼 ∞ 𝑙𝑙 𝑙𝑙 − 𝑏𝑏 𝑏𝑏
𝜙𝜙 = �𝐁𝐁 ⋅ 𝑑𝑑𝐒𝐒 = � � ( + ) 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
𝑆𝑆 4𝜋𝜋 𝑥𝑥=𝑟𝑟 𝑏𝑏=0 𝑥𝑥�𝑥𝑥 2 + (𝑙𝑙 − 𝑏𝑏)2 𝑥𝑥√𝑥𝑥 2 + 𝑏𝑏 2
1 Edward B. Rosa, Bulletin of the Bureau of Standards, vol. 4, no.2 , pp 301-305, 1907.
, which is found to be:
𝜇𝜇0 𝐼𝐼 𝑙𝑙 + √𝑟𝑟 2 + 𝑙𝑙 2 𝑟𝑟 √𝑟𝑟 2 + 𝑙𝑙 2
𝜙𝜙 = 𝑙𝑙[𝑙𝑙𝑙𝑙 + − ]
2𝜋𝜋 𝑟𝑟 𝑙𝑙 𝑙𝑙
From this the external inductance is:
𝜇𝜇0 𝑙𝑙 + √𝑟𝑟 2 + 𝑙𝑙 2 𝑟𝑟 √𝑟𝑟 2 + 𝑙𝑙 2
𝐿𝐿𝑒𝑒𝑒𝑒𝑒𝑒 =
𝑙𝑙[𝑙𝑙𝑙𝑙 + − ]
2𝜋𝜋 𝑟𝑟 𝑙𝑙 𝑙𝑙
And the total inductance would be:
𝜇𝜇0 𝑙𝑙 + √𝑟𝑟 2 + 𝑙𝑙 2 𝑟𝑟 1 √𝑟𝑟 2 + 𝑙𝑙 2
𝐿𝐿 = 𝐿𝐿𝑖𝑖𝑖𝑖𝑖𝑖 + 𝐿𝐿𝑒𝑒𝑒𝑒𝑒𝑒 =
𝑙𝑙[𝑙𝑙𝑙𝑙 + + − ]
2𝜋𝜋 𝑟𝑟 𝑙𝑙 4 𝑙𝑙
For 𝑟𝑟 ≪ 𝑙𝑙, the inductance is roughly:
𝜇𝜇0 2𝑙𝑙 3
𝐿𝐿 ≈ 𝑙𝑙(𝑙𝑙𝑙𝑙 − )
2𝜋𝜋 𝑟𝑟 4
For typical values of 𝑟𝑟 = 12.5𝜇𝜇𝜇𝜇 , and 𝑙𝑙 = 2𝑚𝑚𝑚𝑚, the inductance is found to be about
2𝑙𝑙
2.01nH. Given the logarithmic nature of the term 𝑙𝑙𝑙𝑙 𝑟𝑟 , as a rule of thumb we assign an
inductance of about 1nH/mm for a piece of wire. For the reference, a 1mm long wire
inductance is 0.87nH.
9. In Faraday’s experiment, assume the switch has a resistance of R, and the two coils are
identical with an inductance of L. The battery voltage is VBAT. Find the time-varying current
in the coil. Assuming the iron toroid has a large permeability, find the magnetic flux in the
second coil and estimate the emf read by the galvanometer.
Solution: We use the following circuit model to obtain the current in the primary:
R
VBAT
emf
VBAT i1 L i2
t
