Engineering Electromagnetics
Lecture 5
Dr.-Ing. Erwin Sitompul
President University
http://zitompul.wordpress.com
2 0 1 3
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EEM 5/1
Engineering Electromagnetics
Chapter 4
Energy and Potential
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Chapter 4
Energy and Potential
Energy Expended in Moving a Point Charge in an Electric Field
 The electric field intensity was defined as the force on a unit
test charge at that point where we wish to find the value of the
electric field intensity.
 To move the test charge against the electric field, we have to
exert a force equal and opposite in magnitude to that exerted
by the field. ► We must expend energy or do work.
 To move the charge in the direction of the electric field, our
energy expenditure turns out to be negative. ► We do not do
the work, the field does.
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Chapter 4
Energy and Potential
Energy Expended in Moving a Point Charge in an Electric Field
 Suppose we wish to move a charge Q a distance dL in an
electric field E, the force on Q arising from the electric field is:
FE  Q E
 The component of this force in the direction dL which we must
overcome is:
FEL  FE  a L  Q E  a L
 The force that we apply must be equal and opposite to the force
exerted by the field:
Fappl   Q E  a L
 Differential work done by external source to Q is equal to:
dW   Q E  a L dL   Q E  d L
• If E and L are perpendicular, the
differential work will be zero
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Chapter 4
Energy and Potential
Energy Expended in Moving a Point Charge in an Electric Field
 The work required to move the charge a finite distance is
determined by integration:
W 

fin al
dW
in it
W  Q 
final
E  dL
init
• The path must be specified beforehand
• The charge is assumed to be at rest at both initial
and final positions
• W > 0 means we expend energy or do work
• W < 0 means the field expends energy or do work
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Chapter 4
Energy and Potential
The Line Integral
 The integral expression of previous equation is an
example of a line integral, taking the form of
integral along a prescribed path.
 Without using vector notation,
we should have to write:
W  Q 
fin al
in it
E L dL
• EL: component of E along dL
 The work involved in moving a charge Q from B to A is
approximately:
W   Q ( E L 1  L1  E L 2  L 2 
W   Q (E1   L 1  E 2   L 2 
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 E L 6  L6 )
 E 6  L 6 )
Erwin Sitompul
EEM 5/6
Chapter 4
Energy and Potential
The Line Integral
 If we assume that the electric field is uniform,
E1  E 2 
 E6
W  Q E  (L1  L 2 
 L 6 )
L BA
 Therefore,
W   Q E  L BA
(uniform E)
 Since the summation can be interpreted as a line integral, the
exact result for the uniform field can be obtained as:
A
W  Q  E  dL
B
A
W  Q E   dL
(uniform E )
W   Q E  L BA
(uniform E) • For the case of uniform E, W
does not depend on the particular
path selected along which the
charge is carried
B
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EEM 5/7
Chapter 4
Energy and Potential
The Line Integral
 Example
Given the nonuniform field E = yax + xay +2az, determine the
work expended in carrying 2 C from B(1,0,1) to A(0.8,0.6,1)
along the shorter arc of the circle x2 + y2 = 1, z = 1.
d L  dx a x  dy a y  dz a z • Differential path, rectangularcoordinate
A
W  Q  E  dL
B
A
  Q  ( ya x  xa y  2a z )  ( d xa x  d ya y  d za z )
B
 2
0 .8
1
yd x  2 
0 .6
0
1
xd y  2  2 d z
1
• Circle equation: x 2  y 2  1
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x
1 y
2
y
1 x
2
Erwin Sitompul
EEM 5/8
Chapter 4
Energy and Potential
The Line Integral
W  2
0.8
1
x
 2 
2
1  x dx  2 
0.6
2
1 x 
2
1
1  y dy  2  2 dz
0
1
0.8
sin
1
2

x
1
1
2
y
2
2
1 y 
2
1
0.6
sin
1
2
  0.962 J

a  u du 
2
2
u

y
0
a u 
2
2
2
a
2
2
sin
1
 Example
Redo the example, but use the straight-line path from B to A.
yA  yB
• Line equation: y  y B 
W  2
0.8
 2 
0.8
1
1
ydx  2 
0.6
0
x A  xB
( x  xB )  y  3 x  3
1
xdy  2  2 dz
(  3 x  3) dx  2 
1
0.6
0
(1 
y
) dy  0
3
  0.962 J
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u
a
Chapter 4
Energy and Potential
Differential Length
d L  dx a x  dy a y  dz a z
Rectangular
d L  d  a    d  a   dz a z
Cylindrical
d L  dr a r  rd  a   r sin  d  a 
Spherical
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Chapter 4
Energy and Potential
Work and Path Near an Infinite Line Charge
L
E  Ea 
a
2  0 
d L  d  a    d  a   dz a z
L
W  Q 
final
 Q 
final
2  0  1
init
L
2  0
init
a   1d  a 
d  a   a
 0
W  Q 
final
init
 Q 
b
L
2  0 
L d 
a d a
2  0 
QL
b
 
ln
2  0
a
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a
EEM 5/11
Chapter 4
Energy and Potential
Definition of Potential Difference and Potential
 We already find the expression for the work W done by an
external source in moving a charge Q from one point to another
in an electric field E:
W  Q 
final
E  dL
init
 Potential difference V is defined as the work done by an
external source in moving a unit positive charge from one point
to another in an electric field:
P o ten tial d ifferen ce  V   
fin al
E  dL
in it
 We shall now set an agreement on the direction of movement.
VAB signifies the potential difference between points A and B
and is the work done in moving the unit charge from B (last
named) to A (first named).
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Chapter 4
Energy and Potential
Definition of Potential Difference and Potential
 Potential difference is measured in joules per coulomb (J/C).
However, volt (V) is defined as a more common unit.
 The potential difference between points A and B is:
• VAB is positive if work is done in carrying
the unit positive charge from B to A
A
V AB    E  d L V
B
 From the line-charge example, we found that the work done in
taking a charge Q from ρ = a to ρ = b was:
W  
QL
2  0
ln
b
a
 Or, from ρ = b to ρ = a,
W  
QL
2  0
ln
a

b
QL
2  0
ln
b
a
 Thus, the potential difference between points at ρ = a to
ρ = b is:
V ab 
W
Q

L
2  0
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ln
b
a
Erwin Sitompul
EEM 5/13
Chapter 4
Energy and Potential
Definition of Potential Difference and Potential
 For a point charge, we can find the potential difference
between points A and B at radial distance rA and rB, choosing
an origin at Q:
E  Erar 
d L  dr a r
Q
4  0 r
2
ar
A
V AB    E  d L
B
 
rA
rB
Q
4  0 r
2
dr
 1
1 




4  0  rA
rB 
Q
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• rB > rA  VAB > 0, WAB > 0,
Work expended by the
external source (us)
• rB < rA  VAB < 0, WAB < 0,
Work done by the electric
field
Erwin Sitompul
EEM 5/14
Chapter 4
Energy and Potential
Definition of Potential Difference and Potential
 It is often convenient to speak of potential, or absolute
potential, of a point rather than the potential difference
between two points.
 For this purpose, we must first specify the reference point
which we consider to have zero potential.
 The most universal zero reference point is “ground”, which
means the potential of the surface region of the earth.
 Another widely used reference point is “infinity.”
 For cylindrical coordinate, in discussing a coaxial cable, the
outer conductor is selected as the zero reference for potential.
 If the potential at point A is VA and that at B is VB, then:
V AB  V A  V B
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Chapter 4
Energy and Potential
The Potential Field of a Point Charge
 In previous section we found an expression for the potential
difference between two points located at r = rA and r = rB in the
field of a point charge Q placed at the origin:
V AB
 1
1 



  V A  VB
4  0  rA rB 
Q
V AB   
rA
rB
E r dr
 Any initial and final values of θ or Φ will not affect the answer.
As long as the radial distance between rA and rB is constant,
any complicated path between two points will not change the
results.
 This is because although dL has r, θ, and Φ components, the
electric field E only has the radial r component.
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Erwin Sitompul
EEM 5/16
Chapter 4
Energy and Potential
The Potential Field of a Point Charge
 The potential difference between two points in the field of a
point charge depends only on the distance of each point from
the charge.
 Thus, the simplest way to define a zero reference for potential
in this case is to let V = 0 at infinity.
 As the point r = rB recedes to infinity, the potential at rA
becomes:
V AB  V A  V B
V AB 
V AB 
V AB 
Q
1
4  0 rA
Q
1
4  0 rA
Q


1
4  0 rA
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Q
1
4  0 rB
Q
1
4 0 
 VA
Erwin Sitompul
EEM 5/17
Chapter 4
Energy and Potential
The Potential Field of a Point Charge
 Generally,
V 
Q
4  0 r
 Physically, Q/4πε0r joules of work must be done in carrying
1 coulomb charge from infinity to any point in a distance of r
meters from the charge Q.
 We can also choose any point as a zero reference:
V 
Q
4  0 r
 C1
with C1 may be selected so that V = 0 at any desired value of r.
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EEM 5/18
Chapter 4
Energy and Potential
Equipotential Surface
 Equipotential surface is a surface composed of all those points
having the same value of potential.
 No work is involved in moving a charge around on an
equipotential surface.
 The equipotential surfaces in the potential field of a point
charge are spheres centered at the point charge.
 The equipotential surfaces in the potential field of a line charge
are cylindrical surfaces axed at the line charge.
 The equipotential surfaces in the potential field of a sheet of
charge are surfaces parallel with the sheet of charge.
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EEM 5/19
Chapter 4
Energy and Potential
Homework 5
 D4.2.
 D4.3.
 D4.4.
 D4.5.
 All homework problems from Hayt and Buck, 7th Edition.
 Due: Monday, 13 May 2013.
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EEM 5/20