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Lecture 6:
Electromotive Force; Kirchoff’s
Laws; Redistribution of Charge;
Boundary Conditions for Steady
Current Flow
Lecture 6
1
To study:
 the electromotive force;
 Kirchoff’s laws;
 charge redistribution in a conductor;
boundary conditions for steady current flow.

Lecture 6
2


Steady current flow requires a closed circuit.
Electrostatic fields produced by stationary
charges are conservative. Thus, they cannot by
themselves maintain a steady current flow.
Lecture 6
3
 
increasing
potential

I

The current I must
be zero since the
electrons cannot
gain back the
energy they lose in
traveling through
the resistor.
Lecture 6
4
 

I
+
-

To maintain a
steady current,
there must be an
element in the
circuit wherein
the potential rises
along the
direction of the
current.
Lecture 6
5


For the potential to rise along the direction of
the current, there must be a source which
converts some form of energy to electrical
energy.
Examples of such sources are:




batteries
generators
thermocouples
photo-voltaic cells
Lecture 6
6
• Eemf is the electric field
established by the energy
conversion.
• This field moves positive
charge to the upper plate,
and negative charge to the
lower plate.
• These charges establish an
electrostatic field E.
+++
E
E emf
--In equilibrium:
E emf  E  0
Source is not connected
to external world.
Lecture 6
7
 
I

+
Eemf
-
E
E
At all points in the circuit, we must
have
J
 E total  E emf  E

exists only in battery
Lecture 6
8

Integrate around the circuit in the
direction of current flow
E
total
dl  
C
C
1


 E dl   E

C
emf
J dl
 d l 
C
1

J dl
0
Lecture 6
9

Define the electromotive force (emf) or
“voltage” of the battery as

Vemf   E emf  d l

Lecture 6
10

We also note that
1
l
C  J  d l  A I  RI

Thus, we have the circuit relation
Vemf  RI
Lecture 6
11
Fundamental laws of
classical electromagnetics
Special
cases
Electrostatics
Statics:
Input from
other
disciplines
Maxwell’s
equations
Magnetostatics
Electromagnetic
waves

0
t
Geometric
Optics
Transmission
Line
Theory
Circuit
Theory
Kirchoff’s
Laws
d  
Lecture 6
12

For a closed circuit containing voltages
sources and resistors, we have
V
emf
 I R
• “the algebraic sum of the emfs around a closed circuit
equals the algebraic sum of the voltage drops over the
resistances around the circuit.”
Lecture 6
13


Strictly speaking KVL only applies to
circuits with steady currents (DC).
However, for AC circuits having
dimensions much smaller than a
wavelength, KVL is also approximately
applicable.
Lecture 6
14


Electric charges can neither be created nor
destroyed.
Since current is the flow of charge and charge is
conserved, there must be a relationship
between the current flow out of a region and
the rate of change of the charge contained
within the region.
Lecture 6
15

Consider a
volume V
bounded by a
closed surface S in
a homogeneous
medium of
permittivity e and
conductivity 
containing charge
density qev.
S
qev
V
ds
Lecture 6
16

The net current
leaving V through
S must be equal to
the time rate of
decrease of the
total charge
within V, i.e.,
S
qev
V
dQenc
I 
dt
ds
Lecture 6
17

The net current leaving the region is given by
I   J ds
S

The total charge enclosed within the region is
given by
Q   qev dv
V
Lecture 6
18

Hence, we have
d
S J  d s   dt V qev dv
net outflow
of current
net rate of
decrease of
total charge
Lecture 6
19

Using the divergence theorem, we have
 J  d s     J dv
S

V
We also have
qev
d
qev dv  
dv

dt V
t
V
Becomes a
partial
derivative when
moved inside of
the integral
because qev is a
function of
position as well
as time.
Lecture 6
20


Thus,

V   J dv  V t dv  0
Since the above relation must be true for any
and all regions, we have

J 
0
t
Continuity
Equation
Lecture 6
21

For steady currents,

Thus,

0
t
J  0
J is a solenoidal vector field.
Lecture 6
22

Ohm’s law in a conducting medium states
J  E

For a homogeneous medium
  J    E  0    E  0
But from Gauss’s law,
qev
E 
e

Therefore, the volume charge density, ,
must be zero in a homogeneous conducting
Lecture 6
medium
23

Since J is solenoidal,
we must have
S
J

d
s

0

S


In a circuit, steady
current flows in
wires.
Consider a “node” in
a circuit.
Lecture 6
24

We have for a node in a circuit
I 0
• “the algebraic sum of all currents leaving a
junction must be zero.”
Lecture 6
25


Strictly speaking KCL only applies to
circuits with steady currents (DC).
However, for AC circuits having
dimensions much smaller than a
wavelength, KCL is also approximately
applicable.
Lecture 6
26

Charges introduced into the interior of
an isolated conductor migrate to the
conductor surface and redistribute
themselves in such a way that the
following conditions are met:




E = 0 within the conductor
Et = 0 just outside the conductor
qev = 0 within the conductor
qes  0 on the surface of the conductor
Lecture 6
27


We can derive the differential equation
governing the redistribution of charge
from Gauss’s law in differential form and
the continuity equation.
From Gauss’s law for the electric field,
we have
  D  qev
qev
 E 
e

   J  qev
e
Lecture 6
28


From the continuity equation, we have
Combining the two equations, we obtain

J  
t
Describes the time
 r , t  
  r , t   0
t
e
evolution of the
charge density at a
given location.
Lecture 6
29

The solution to the DE is given by
 r , t    0 r  e
 t /  r 
Initial charge distribution at t = 0
where r = e/ is the time constant of the
process called the relaxation time.
Lecture 6
30




The initial charge distribution at any point
in the bulk of the conductor decays
exponentially to zero with a time constant r.
At the same time, surface charge is building
up on the surface of the conductor.
The relaxation time decreases with increasing
conductivity.
For a good conductor, the time required for
the charge to decay to zero at any point in
the bulk of the conductor (and to build up
on the surface of the conductor) is very
small.
Lecture 6
31

copper
r

H 2O
r

amber
r

mica
r

quartz
r
 1.5 10
19
s
5
 10 s
 4 10 s
3
 10 to 20 hrs
 50 days
Lecture 6
32


The concept of relaxation time is also
used to determine the electrical
nature (conductor or insulator) of
materials at a given frequency.
A material is considered to be a
good conductor if
1
 r  T 
f
  r f  1
Lecture 6
33

A material is considered to be a good insulator
if
1
 r  T 
f

  r f  1
A good conductor is a material with a
relaxation time such that any free charges
deposited within its bulk migrate to its surface
long before a period of the wave has passed.
Lecture 6
34


The behavior of current flow across
the interface between two different
materials is governed by boundary
conditions.
The boundary conditions for current
flow are obtained from the integral
forms of the basic equations governing
current flow.
Lecture 6
35
ân
e1 ,  1
e 2 , 2
Lecture 6
36

The governing equations for steady electric
current (in a conductor) are:
 J ds  0
S
E

d
l

0



C
C
J

dl  0
Lecture 6
37

The normal component of a solenoidal
vector field is continuous across a
material interface:
J 1n  J 2 n

The tangential component of a
conservative vector field is continuous
across a material interface:
J 1t
1

J 2t
2
38
Lecture 6
 0
J=0
 0
J
Lecture 6
39



The current in the conductor must flow
tangential to the boundary surface.
The tangential component of the electric
field must be continuous across the
interface.
The normal component of the electric
field must be zero at the boundary inside
the conductor, but not in the dielectric.
Thus, there will be a buildup of surface
charge at the interface.
Lecture 6
40
+++++
no current flow  Et = 0
E
----+++++
current flow  En >> Et
I
----Lecture 6
41

The current bends as it cross the interface
between two conductors
1
 2  1
1
2
Lecture 6
42


The angles are related by
Suppose medium 1 is a good conductor
and medium 2 is a good insulator (i.e., 1
>> 2). Then
. In other words, the
current enters medium 2 at nearly right
 2 to
 0the boundary. This result is
angles
consistent with the fact that the electric
field in medium 2 should have a
vanishingly small tangential component
at the interface.
Lecture 6
43

In general, there is a buildup of surface charge
at the interface between two conductors.
J1n  J 2 n  J n   1 E1n   2 E2 n
 s  D1n  D2 n  e1 E1n  e 2 E2 n

 e1 e 2 
1 
 E1n  J n   
  e1  e 2
2 

 1  2 
 J n  r1   r 2 
 Only when the
relaxation times of
the two conductors
are equal is there no
buildup of surface
charge at the
interface.
Lecture 6
44