Transcript Displacement Current and the Generalized Ampere`s Law

Displacement Current and the
Generalized Ampere’s Law
AP Physics C
Montwood High School
R. Casao
• Charges in motion, or currents, produce
magnetic fields.
• When a current-carrying conductor has high
symmetry, we can determine the magnetic field
using Ampere’s law:
 B  ds  μ  I
o
enclosed
– where the line integral is over any closed path
through which the conduction current passes.
– Conduction current is the current carried by the
wire.
dQ
– The conduction current is defined by:
I
dt
• Ampere’s law in this form is only valid if the
conduction current is constant over time. This
means that any electric fields present are constant
as well.
• Maxwell recognized this limitation of Ampere’s
law and modified the law to include electric
fields that change over time.
• Consider a charging capacitor:
– The current I is decreasing over time as the
magnitude of the charge on the capacitor and the
electric field between the capacitor plates increases.
– No conduction current
passes between the
capacitor plates.
• Consider the two
Amperian surfaces S1 and
S2 bounded by the same
path P.
– Ampere’s law says that the
line integral of B•ds
around this path must
equal µo·I, where I is the
total current through any
surface bounded by the
path P.
• When the path P bounds
S1, the result of the
integral is µo·I since the
conduction current passes
through S1.
• When the path bounds S2,
the result of the integral is
zero since no conduction
current passes through S2.
• Current passes through S1
but does not pass through
S2.
• Maxwell solved this problem by adding a
displacement current Id to the right side of

Ampere’s law: 
 B  ds  μ  I
o
enclosed
 Id 
• Displacement current Id is proportional to the
rate of change of the electric flux E.
dΦ E
Id  ε o 
dt
• Electric flux E is defined as:


Φ E   E  dA
• As the capacitor is being charged (or discharged),
the changing electric field between the plates can
be considered as a current that bridges the
discontinuity in the conduction current.
• Adding the displacement current to the right side
of Ampere’s law allows some combination of
conduction current and displacement current to
pass through surfaces S1 and S2.
• Ampere-Maxwell law:


 B  d s  μ o  I conduction  I displacement 
 
dΦ E
 B  d s  μ o I conduction  μ o  ε o 
dt
• The electric flux through S2 is:
  

Φ E   E  dA  E  A
– E is the uniform electric field between the plates.
– A is the area of the plates.
• The conduction current passes through S1.
• If the charge on each plate at any instant is Q,

then:

Q
εo  A
V
C
C
E
V
d
d


εo  A
Q
Q


E
d
Ed
εo  A
• The electric flux through S2 is:
 
Q  Q
 A 
ΦE  E  A 
εo
εo  A
• The displacement current Id through S2 is:
Q


d

εo 
dΦ E

Id  ε o 
 εo 
dt
dt
1 dQ dQ
Id  ε o  

ε o dt
dt
• The conduction current is equal to the
displacement current: I = Id
• Magnetic fields are produced by both conduction
currents and changing electric fields.
Displacement Current in a Capacitor
• An alternating current (AC) voltage is applied
directly across an 8 F capacitor. The
frequency of the AC source is 3 kHz and the
voltage amplitude (Vmax) is 30 V. Determine
the displacement current between the plates of
the capacitor.
• Angular frequency of the oscillation is:
 = 2·  ·f
• Voltage as a function of time:
V = Vmax·sin( ·t)
• Charge on the capacitor is Q = C·V
dQ d(C  V)
dV
Id 

 C
dt
dt
dt
dV dVmax  sin ω  t 
dsin ω  t 

 Vmax 
dt
dt
dt
dV
 Vmax  ω  cosω  t 
dt
I d  C  Vmax  ω  cosω  t 
• Substituting:
ω  2  π  f  2  π  3000Hz  6000  π Hz
I d  C  Vmax  ω  cos6000  π Hz  t 
I d  8 x 106 F  30 V  6000  π Hz  cos6000  π Hz  t 
I d  4.5239 A  cos6000  π Hz  t 
• The displacement current graphs as a sine wave
with a maximum value of 4.5239 A.