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Engineering Electromagnetics
Lecture 6
Dr.-Ing. Erwin Sitompul
President University
http://zitompul.wordpress.com
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President University
Erwin Sitompul
EEM 6/1
Chapter 4
Energy and Potential
The Potential Field of a System of Charges: Conservative Property
 We will now prove, that for a system of charges, the potential is
also independent of the path taken.
 Continuing the discussion, the potential field at the point r due
to a single point charge Q1 located at r1 is given by:
V (r) 
Q1
4 0 r  r1
 The field is linear with respect to charge so that superposition is
applicable. Thus, the potential arising from n point charges is:
Q1
Q2
V (r) 


4 0 r  r1 4 0 r  r2
Qn

4 0 r  rn
n
Qm
V (r )  
m 1 4 0 r  rm
President University
Erwin Sitompul
EEM 6/2
Chapter 4
Energy and Potential
The Potential Field of a System of Charges: Conservative Property
 If each point charge is now represented as a small element of
continuous volume charge distribution ρvΔv, then:
v (r1 )v1
v (r2 )v2
v (rn )vn
V (r) 

 
4 0 r  r1 4 0 r  r2
4 0 r  rn
 As the number of elements approach infinity, we obtain the
integral expression:
v (r)dv
V (r)  
vol 4 r  r
0
 If the charge distribution takes from of a line charge or a
surface charge,
 L (r)dL
V (r)  
4 0 r  r
 (r)dS 
V (r)   S
S 4 r  r
0
President University
Erwin Sitompul
EEM 6/3
Chapter 4
Energy and Potential
The Potential Field of a System of Charges: Conservative Property
 As illustration, let us find V on the z axis for a uniform line
charge ρL in the form of a ring, ρ = a, in the z = 0 plane.
 L (r)dL
V (r)  
4 0 r  r

2
0

 L ad 
4 0 a 2  z 2
La
2 0 a 2  z 2
• The potential arising from point charges
or continuous charge distribution can
be seen as the summation of potential
arising from each charge or each
differential charge.
• It is independent of the path chosen.
President University
Erwin Sitompul
EEM 6/4
Chapter 4
Energy and Potential
The Potential Field of a System of Charges: Conservative Property
 With zero reference at ∞, the expression for potential can be
taken generally as:
A
VA   E  dL

 Or, for potential difference:
A
VAB  VA  VB   E  dL
B
 Both expressions above are not
dependent on the path chosen for
the line integral, regardless of the
source of the E field.
• Potential conservation
in a simple dc-circuit
problem in the form of
Kirchhoff’s voltage law
 For static fields, no work is done in
carrying the unit charge around any
closed path.
 E  dL  0
President University
Erwin Sitompul
EEM 6/5
Chapter 4
Energy and Potential
Potential Gradient
 We have discussed two methods of determining potential:
directly from the electric field intensity by means of a line
integral, or from the basic charge distribution itself by a volume
integral.
 In practical problems, however, we rarely know E or ρv.
 Preliminary information is much more likely to consist a
description of two equipotential surface, and the goal is to find
the electric field intensity.
President University
Erwin Sitompul
EEM 6/6
Chapter 4
Energy and Potential
Potential Gradient
 The general line-integral
relationship between V and E is:
V    E  dL
dV  E  dL
 For a very short element of length
ΔL, E is essentially constant:
V
E  L
V
V
L
EL cos
 E cos 
 Assuming a conservative field, for a given reference and
starting point, the result of the integration is a function of the
end point (x,y,z). We may pass to the limit and obtain:
dV
  E cos 
dL
President University
Erwin Sitompul
EEM 6/7
Chapter 4
Energy and Potential
Potential Gradient
 From the last equation, the maximum positive increment of
potential, Δvmax, will occur when cosθ = –1, or ΔL points in the
direction opposite to E.
dV
dL
E
max
 We can now conclude two characteristics of the relationship
between E and V at any point:
1. The magnitude of E is given by the maximum value of the
rate of change of V with distance L.
2. This maximum value of V is obtained when the direction of
the distance increment is opposite to E.
President University
Erwin Sitompul
EEM 6/8
Chapter 4
Energy and Potential
Potential Gradient
 For the equipotential surfaces below,
find the direction of E at P.
E
dV
dL
,
max
  180
President University
Erwin Sitompul
EEM 6/9
Chapter 4
Energy and Potential
Potential Gradient
 Since the potential field information is more likely to be
determined first, let us describe the direction of ΔL (which leads
to a maximum increase in potential) in term of potential field.
 Let aN be a unit vector normal to the equipotential surface and
directed toward the higher potential.
 The electric field intensity is then expressed in terms of the
potential as:
dV
E=
dL
aN
max
 The maximum magnitude occurs when ΔL is in the aN direction.
Thus we define dN as incremental length in aN direction,
dV
dL

max
E=
dV
dN
dV
aN
dN
President University
Erwin Sitompul
EEM 6/10
Chapter 4
Energy and Potential
Potential Gradient
 We know that the mathematical operation to find the rate of
change in a certain direction is called gradient.
 Now, the gradient of a scalar field T is defined as:
Gradient of T  grad T 
dT
aN
dN
 Using the new term,
dV
E=
a N = grad V
dN
President University
Erwin Sitompul
EEM 6/11
Chapter 4
Energy and Potential
Potential Gradient
 Since V is a function of x, y, and z, the total differential is:
V
V
V
dV 
dx 
dy 
dz
x
y
z
 But also,
dV  E  dL   Ex dx  E y dy  Ez dz
 Both expression are true for any dx, dy, and dz. Thus:
V
x
V
Ey  
y
V
Ez  
z
Ex  
 V
V
V 
E  
ax 
ay 
az 
y
z 
 x
grad V 
V
V
V
ax 
ay 
az
x
y
z
 Note: Gradient of a scalar is a vector.
President University
Erwin Sitompul
EEM 6/12
Chapter 4
Energy and Potential
Potential Gradient
 Introducing the vector operator for gradient:



  ax  a y  az
x
y
z
We now can relate E and V as:
E  V
V 
V
V
V
ax 
ay 
az
x
y
z
Rectangular
V
1 V
V
V 
a 
a 
az

 
z
V 
V
1 V
1 V
ar 
a 
a
r
r 
r sin  
President University
Erwin Sitompul
Cylindrical
Spherical
EEM 6/13
Chapter 4
Energy and Potential
Potential Gradient
 Example
Given the potential field, V = 2x2y–5z, and a point P(–4,3,6),
find V, E, direction of E, D, and ρv.
VP  2(4) 2 (3)  5(6)  66 V
 V
V
V 
E  V   
ax 
ay 
a z   4 xya x  2 x2a y  5a z
y
z 
 x
EP  4(4)(3)a x  2(4)2 a y  5a z  48a x  32a y  5a z V m
aE,P 
EP
EP
DP   0E P  425a x  283.3a y  44.27a z pC m3
v  div D  div  0E  (8.854 1012 )(4 y)  35.42 y pC m3
At P, v  35.42(3) pC m3  106.26 pC m3
President University
Erwin Sitompul
EEM 6/14
Chapter 4
Energy and Potential
The Dipole
 The dipole fields form the basis for the behavior of dielectric
materials in electric field.
 The dipole will be discussed now and will serve as an
illustration about the importance of the potential concept
presented previously.
 An electric dipole, or simply a dipole, is the name given to two
point charges of equal magnitude and opposite sign, separated
by a distance which is small compared to the distance to the
point P at which we want to know the electric and potential
fields.
President University
Erwin Sitompul
EEM 6/15
Chapter 4
Energy and Potential
The Dipole
 The distant point P is described by the spherical coordinates
r, θ, Φ = 90°.
 The positive and negative point charges have separation d and
described in rectangular coordinates (0,0, 0.5d) and
(0,0,–0.5d).
President University
Erwin Sitompul
EEM 6/16
Chapter 4
Energy and Potential
The Dipole
 The total potential at P can be written as:
Q 1 1 
Q R2  R1
V




4 0  R1 R2  4 0 R1 R2
 The plane z = 0 is the locus of points for which R1 = R2
► The potential there is zero (as also all points at ∞).
President University
Erwin Sitompul
EEM 6/17
Chapter 4
Energy and Potential
The Dipole
 For a distant point, R1 ≈ R2 ≈ r, R2–R1 ≈ dcosθ
Qd cos 
V
4 0 r 2
 Using the gradient in spherical coordinates,
E  V
 V
1 V
1 V 
 
ar 
a 
a 
r 
r sin   
 r
 Qd cos 
Qd sin  
E   
ar 
a 
3
3
4 0 r
 2 0 r

E
Qd
2 cos  a r  sin  a 
3 
4 0 r
President University
Erwin Sitompul
EEM 6/18
Chapter 4
Energy and Potential
The Dipole
 To obtain a plot of the potential
field, we choose Qd/(4πε0) = 1
and thus cosθ = Vr2.
 The colored lines in the figure
below indicate equipotentials for
V = 0, +0.2, +0.4, +0.6, +0.8,
and +1.
r = 2.236
r = 1.880
Plane at
zero potential
45°
Qd cos 
V
4 0 r 2
President University
Erwin Sitompul
EEM 6/19
Chapter 4
Energy and Potential
The Dipole
 The potential field of the dipole may be simplified by making
use of the dipole moment.
 If the vector length directed from –Q to +Q is identified as d,
then the dipole moment is defined as Qd and is assigned the
symbol p.
p  Qd
 Since dar = dcosθ , we then have:
p  ar
V
4 0 r 2
r  r
V
p
2
r  r
4 0 r  r
1
President University
• Dipole charges: V
1
,E
2
r
• Point charge:
1
,E
r
Erwin Sitompul
V
EEM 6/20
1
r3
1
r2
Chapter 4
Energy and Potential
Exercise Problems
1. A charge in amount of 13.33 nC is uniformly distributed in a circular disk
form, with the radius of 2 m. Determine the potential at a point on the axis,
2 m from the disk. Compare this potential with that which results if all the
charge is at the center of the disk.
(Sch.S62.E3)
Answer: 49.7 V, 60 V.
2. For the point P(3,60°,2) in cylindrical coordinates and the potential field
V = 10(ρ +1)z2cosφ V in free space, find at P: (a) V; (b) E; (c) D; (d) ρv; (e)
dV/dN; (f) aN.
(Hay.E5.S112.23)
Answer: (a) 80 V; (b) –20aρ + 46.2aφ – 80az V/m; (c) –177.1aρ + 409aφ –
708az pC/m2; (d) –334 pC/m3;(e) 94.5 V/m; (f) 0.212aρ – 0.489aφ +0.846az
3. Two opposite charges are located on the z-axis and centered on the origin,
configuring as an electric dipole is located at origin. The distance between
the two charges, each with magnitude of 1 nC, is given by 0.1 nm. The
electric potential at A(0,1,1) nm is known to be 2 V. Find out the electric
potential at B(1,1,1) nm. Hint: Do not assume that A and B are distant
points. Determine E due to both charges first, then calculate the potential
difference.
(EEM.Pur5.N4)
Answer: 1.855 V.
President University
Erwin Sitompul
EEM 6/21