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Today’s agenda:
Induced Electric Fields.
You must understand how a changing magnetic flux induces an electric field, and be able
to calculate induced electric fields.
Eddy Currents.
You must understand how induced electric fields give rise to circulating currents called
“eddy currents.”
Displacement Current and Maxwell’s Equations.
Displacement currents explain how current can flow “through” a capacitor, and how a timevarying electric field can induce a magnetic field.
Back emf.
A current in a coil of wire produces an emf that opposes the original current.
Displacement Current
Apply Ampere’s Law to a
charging capacitor.
 B  ds = μ0IC
ds
IC
+
+q
-q
IC
The shape of the surface used for Ampere’s Law shouldn’t
matter, as long as the “path” is the same.
Imagine a soup bowl
surface, with the + plate
resting near the bottom of
the bowl.
Apply Ampere’s Law to the
charging capacitor.
ds
IC
+
+q
-q
IC
 B  ds = 0
The integral is zero because no current passes through the
“bowl.”
 B  ds = μ I
0 C
 B  ds = 0
Hold it right there pal! You can’t have it both ways. Which is it
that you want? (The equation on the right is actually incorrect, and the equation on the left is incomplete.)
As the capacitor charges, the electric field between the plates
changes.
A

q = CV =  ε 0  Ed 
d

= ε 0EA = ε 0E
ds
IC
As the charge and electric field
change, the electric flux changes.
E
+
+q
-q
IC
dq d
d
=  ε 0E  = ε 0  E 
dt dt
dt
This term has units of current.
 Is the dielectric constant of the medium in between the capacitor plates. In the diagram, with an air-filled capacitor,  = 1.
We define the displacement current to be
d
ID = ε 0  E  .
dt
The changing electric flux
through the “bowl” surface
is equivalent to the current
IC through the flat surface.
ε = ε 0
ds
IC
E
+
+q
-q
IC
The generalized (“always correct”) form of Ampere’s Law is
then
dE
 B  ds = μ0  IC  ID encl = μ0Iencl  μ0ε dt .
Magnetic fields are produced by both conduction currents and
time varying electric fields.
The “stuff” inside the gray boxes serves as your official starting
equation for the displacement current ID.
dE
 B  ds = μ0  IC  ID encl = μ0Iencl  μ0ε dt .
 is the relative dielectric constant; not emf.
In a vacuum, replace  with 0.
ID
Why “displacement?” If you put an insulator in between the plates of the capacitor, the atoms of the insulator are “stretched”
because the electric field makes the protons “want” to go one way and the electrons the other. The process of “stretching” the
atom involves displacement of charge, and therefore a current.
Homework Hints
d E
Displacement current: ID = ε
dt
where ε = ε 0
 is the relative dielectric constant; not emf
Homework Hints
For problem 9.45a, recall that
I
E = J =
A
For problem 9.45c, displacement current density is calculated
the same as conventional current density
ID
JD =
A
Not applicable Spring 2015.
The Big Picture
You have now learned Gauss’s Law for both electricity and
magnetism, Faraday’s Law of Induction,
Induction…and the generalized
form of Ampere’s Law:
q enclosed
 E  dA  o
d B
 E  ds   dt
 B  dA  0
dΦ E
 B  ds=μ 0 Iencl +μ 0ε 0 dt .
These equations can also be written in differential form:

E 
0
 B  0
dB
×E=dt
1 dE
  B= 2
+μ 0 J
c dt
Congratulations! It is my great honor
The Missouri
S&T Society
of qualified
PhysicstoStudent
to pronounce
you fully
wear… T-Shirt!
…with all the rights and privileges thereof.*
This will not be tested on the exam.
*Some day I might figure out what these rights and privileges are. Contact me if you want to buy an SPS t-shirt for the member price of $10.