Transcript Slide 1
Engineering Electromagnetics
Lecture 5
Dr.-Ing. Erwin Sitompul
President University
http://zitompul.wordpress.com
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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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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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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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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( EL1L1 EL2L2
W Q(E1 L1 E2 L2
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EL6L6 )
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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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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EEM 5/8
Chapter 4
Energy and Potential
Differential Length
dL dxax dya y dzaz
Rectangular
dL d a da dzaz
Cylindrical
dL drar rd a r sin da
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 da dzaz
E E a
L
a 1da
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
QL b
ln
2 0 a
W Q
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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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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:
QL b
W
ln
2 0 a
Or, from ρ = b to ρ = a,
QL a
QL 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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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