HW5_EO

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ECE 339
Electromagnetics
Homework 5
Due Lecture #18
(100 points)
1. (Gauss’s Law/10 points) You are in a car that is struck by lightning. The charge (about 10-20
Coulombs) cannot leak to ground, so spreads itself evenly over the metal surface of the car.
(a) Will the electric field inside by high or low? Will you be in danger? Explain.
(b) As you look out the window of your car, you notice a purplish glow at a corner on the car’s
surface. This is called corona and is a region of very intense electric field - intense enough
to ionize the air in its vicinity and cause the air to glow. If the dielectric breakdown field of
air is 3 × 106 Volts per meter, and the radius of curvature where the corona is observed is
approximately 10 cm, calculate the surface charge needed for corona to occur. (Hint: The
electric field at the surface of a sphere of radius r with a surface charge density ρs Coulombs
per square meter is
E =
Q
4π�0r2
where Q = 4πr2ρs.)
(c) Explain why corona occurs at a corner. This is the same reason that corona can sometimes
be seen at the end of a lightning rod.
2. (Dielectrics, Flux Density/10 points) A positive point charge Q is at the center of a spherical
dielectric shell of an inner radius a and an outer radius b. The dielectric constant of the shell is
�r, so the shell has a permittivity � = �r�0. Determine the electric field intensity ~E Volts/meter,
the Electric Flux Density ~D Coulombs/m2, and the Polarization Vector ~P Coulombs/m2 in all
regions. Plot ~D and ~E in all three regions.
3. (Laplace’s Equation/10 points) Show that in charge free space (ρv = 0), for an inhomogene-
nous material (i.e., � = �(x, y, z)) that Laplace’s equation does not hold, and the correct equation
is
�∇2V +∇V · ∇� = 0
A useful identity is (where a is a scalar field and ~B is a vector field)
∇ · (a ~B) = a∇ · ~B + ~B · ∇a
4. (Polarization/10 points) In class we showed that a polarized, characterized by a polarization
vector ~P , is equivalent to a volume charge density ρpv = −∇· ~P and an equivalent surface charge
density ρs = ~P · d~S where d~S is the elemental surface vector. We then wrote Gauss’s Law as
∇ · ~E = 1
�0
(ρv + ρpv)
where ρv is the true volume charge density. We were then able to define the Flux Density vector
~D = �0 ~E + ~P . Why is the equivalent surface charge density ρs not included in this derivation?
5. (Electric Potential and Energy/10 points) A 15 µF capacitor is charged to 15 Volts.
a. What is the charge on each plate of the capacitor?
b. How many electrons are on the negative plate of the capacitor assuming the charge on an
electron is q = 1.602× 10−19 Coulombs?
c. What is the potential energy of all the electrons on the negative plate relative to the positive
plate?
d. What is the potential energy of a single electron relative to the other plate of the capacitor?
e. If all the potential energy of one charge is converted to kinetic energy by WE = 12mev
2
where the mass of an electron is me = 9.11× 10−31 kg, what would the velocity be?
6. (Electric Potential and Fields/20 points) The potential of a point charge Q is
V =
Q
4π�0r
Two point charges of equal and opposite sign (+Q and −Q) are separated by a distance L,
forming an electric dipole of dipole moment QL Coul-m. The field is measured at a point (the
Field point) located a distance r from the center of the dipole.
(a) Assuming that r � L show that the distance to the field point from charge Q is r1 ≈
r − L2 cos θ and from −Q is r2 ≈ r +
L
2 cos θ.
(b) Show that the potential for distances r � L is
V =
QL cos θ
4π�0r2
(c) By taking the gradient in spherical coordinates, show that the field of an electric dipole is
~E = âr
QL cos θ
2π�0r3
+ âθ
QL sin θ
4π�0r3
Note that this results only apply in the case that r � L. Also note that the potential of the
electric dipole varies as 1/r2 and the field of the electric dipole varies as 1/r3.
7. (Electric Potential/20 points) Calculate the electric Field ~E at a distance r from an infinite
line charge, density ρl Coulombs per meter by first calculating the electric potential V . Show
that the field is identical to what we derived in class
~E =
ρl
2π�0r
r̂
Hint: You will find that difficulties will arise when integrating from −∞ to ∞. Try this:
Calculate the potential and then the electric field for a segment of line charge L meters long (i.e.,
integrate from −L/2 to +L/2). Once the potential and electric field are found, take the limit of
the field as L → ∞. A difficulty that often arises when dealing with potentials is that infinities
sometimes show up that must be dealt with carefully. Be careful with the math - a fairly involved
derivative is required.
8. (Capacitance, Electric Energy/10 points)
(a) Calculate the capacitance of the Earth. Hint: Assume a radius of the Earth of 6378 km. If
a charge Q is spread uniformly over the Earth’s surface, what is the electrostatic potential
V surrounding the Earth? The definition of capacitance C comes from Q = CV .
(b) In the last homework we learned about an ever-present electric field called the “Fair Weather”
electric field that points radially outward from the Earth’s surface. At the Earth’s surface
this field has a magnitude of about ~E = 300 Volts/meter. In Homework #4 we used
Gauss’s Law to find that the total charge Q on the surface of the Earth is approximately
Q = 1.358× 106 Coulombs. From
~E = âR
Q
4π�0R2
Calculate the total energy stored in this field by integrating
WE =
�0
2
∫ 2π
0
∫ π
0
∫ ∞
rEarth
E2 R2 sin θ dR dθ dφ