Transcript Slide 1

Engineering Electromagnetics
Lecture 5
Dr.-Ing. Erwin Sitompul
President University
http://zitompul.wordpress.com
President University
Erwin Sitompul
EEM 5/1
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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Erwin Sitompul
EEM 5/2
Chapter 4
Energy and Potential
Energy Expended in Moving a Point Charge in an Electric Field
 To move a charge Q a distance dL in an electric field E, the
force on Q arising from the electric field is:
FE  QE
 The component of this force in the direction dL is:
FEL  FE  a L  QE  a L
 The force that we apply must be equal and opposite to the
force exerted by the field:
Fappl  QE  aL
 Differential work done by external source to Q is equal to:
dW  QE  aL dL  QE  dL
• If E and L are perpendicular, the
differential work will be zero
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Erwin Sitompul
EEM 5/3
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 
final
dW
init
W  Q
final
init
E  dL
• 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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Erwin Sitompul
EEM 5/4
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 
final
init
EL dL
• EL: component of E along dL
 The work involved in moving a charge Q from B to A is
approximately:
W  Q( EL1L1  EL2L2 
W  Q(E1  L1  E2  L2 
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 EL6L6 )
 E6  L6 )
Erwin Sitompul
EEM 5/5
Chapter 4
Energy and Potential
The Line Integral
 If we assume that the electric field is uniform,
E1  E2   E6
W  QE  (L1  L2   L6 )
L BA
 Therefore,
W  QE  LBA
(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  QE   dL
(uniform E)
W  QE  LBA
(uniform E) • For the case of uniform E, W
B
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does not depend on the particular
path selected along which the
charge is carried
Erwin Sitompul
EEM 5/6
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.
dL  dxax  dya y  dzaz • Differential path, rectangularcoordinate
A
W  Q  E  dL
B
A
 Q ( ya x  xa y  2a z )  (dxa x  dya y  dza z )
 2
B
0.8
1
0.6
1
0
1
ydx  2 xdy  2 2dz
• Circle equation: x2  y 2  1
x  1 y2
y  1  x2
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Erwin Sitompul
EEM 5/7
Chapter 4
Energy and Potential
The Line Integral
W  2
0.8
1
1  x dx  2
2
0.6
1
1  y dy  2 2dz
2
0
1
0.8
0.6
1
1
x

y

 2  1  x 2  sin 1 x   2  1  y 2  sin 1 y 
2
2
2
1
2
0
 0.962 J
u

a2
u
a  u du 
a  u  sin 1
2
2
a
2
2
2
2
 Example
Redo the example, but use the straight-line path from B to A.
y A  yB
( x  xB )  y  3x  3
xA  xB
• Line equation: y  yB 
W  2
0.8
1
0.6
1
0
1
ydx  2 xdy  2 2dz
0.8
 2 (3x  3)dx  2
1
0.6
0
y
(1  )dy  0
3
 0.962 J
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Erwin Sitompul
EEM 5/8
Chapter 4
Energy and Potential
Differential Length
dL  dxax  dya y  dzaz
Rectangular
dL  d a   da  dzaz
Cylindrical
dL  drar  rd a  r sin  da
Spherical
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Erwin Sitompul
EEM 5/9
Chapter 4
Energy and Potential
Work and Path Near an Infinite Line Charge
L
a
2 0 
dL  d a   da  dzaz
E  E a  
L
a   1da
init 2 
0 1
final 
L
 Q 
d a   a
init 2
0
W  Q 
final
0
L
a  d a 
init 2  
0
b 
L d
 Q 
a 2
0 
QL b

ln
2 0 a
W  Q 
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Erwin Sitompul
final
EEM 5/10
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
init
E  dL
 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:
Potential difference  V  
final
init
E  dL
 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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Erwin Sitompul
EEM 5/11
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:
A
VAB   E  dL V
B
• VAB is positive if work is done in carrying
the positive charge from B to A
 From the line-charge example, we found that the work done in
taking a charge Q from ρ = a to ρ = b was:
QL b
W 
ln
2 0 a
 Or, from ρ = b to ρ = a,
QL a
QL b
W 
ln 
ln
2 0 b
2 0 a
 Thus, the potential difference between points at ρ = a to
ρ = b is:

W
b
Vab   L ln
Q 2 0 a
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Erwin Sitompul
EEM 5/12
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  Er ar 
dL  drar
Q
4 0 r
2
ar
A
VAB   E  dL
B
 
rA
Q
dr
4 0 r
Q 1 1

  
4 0  rA rB 
rB
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2
• 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 5/13
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:
VAB  VA  VB
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Erwin Sitompul
EEM 5/14
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:
VAB
Q 1 1

    VA  VB
4 0  rA rB 
rA
VAB   Er dr
rB
 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 5/15
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:
VAB  VA  VB
Q 1
Q 1
VAB 

4 0 rA 4 0 rB
Q 1
Q 1
VAB 

4 0 rA 4 0 
VAB 
Q
1
 VA
4 0 rA
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Erwin Sitompul
EEM 5/16
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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Erwin Sitompul
EEM 5/17
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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Erwin Sitompul
EEM 5/18
Chapter 4
Energy and Potential
Homework 5
 D4.2.
 D4.3.
 D4.4.
 D4.5. (Bonus Question, + 20 points if correctly made)
 All homework problems from Hayt and Buck, 7th Edition.
 Deadline: 15 May 2012, at 08:00.
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Erwin Sitompul
EEM 5/19