president universityerwin sitompuleem 6/1 lecture 6 engineering electromagnetics dr.-ing. erwin...

President University Erwin Sitompul EEM 6/1

Lecture 6

Engineering Electromagnetics

Dr.-Ing. Erwin SitompulPresident University

http://zitompul.wordpress.com

2 0 1 3


The Potential Field of a System of Charges: Conservative PropertyChapter 4 Energy and Potential

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:

The field is linear with respect to charge so that superposition is applicable. Thus, the potential arising from n point charges is:

1 2

0 1 0 2 0

( )4 4 4

n

n

QQ QV

rr r r r r r

1 0

( )4

nm

m m

QV

rr r

1

0 1

( )4

QV

rr r


Chapter 4 Energy and Potential

vol0

( )( )

4v dv

V

rrr r

The Potential Field of a System of Charges: Conservative Property

As the number of elements approach infinity, we obtain the integral expression:

If each point charge is now represented as a small element of continuous volume charge distribution ρvΔv, then:

1 1 2 2

0 1 0 2 0

( ) ( ) ( )( )

4 4 4v v v n n

n

v v vV

r r rrr r r r r r

If the charge distribution takes from of a line charge or a surface charge,

0

( )( )4

L dLV

rrr r

0

( )( )4

SS

dSV

rrr r




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.

2

2 2004Lad

a z

0

( )( )4

L dLV

rrr r

2 202

La

a z

•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.



A

AV d

E L

A

AB A B BV V V d E L

0d E L


With zero reference at ∞, the expression for potential can be taken generally as:

Or, for potential difference:

Both expressions above are not dependent on the path chosen for the line integral, regardless of the source of the E field.

For static fields, no work is done in carrying the unit charge around any closed path.

•Potential conservation in a simple dc-circuit problem in the form of Kirchhoff’s voltage law



Potential GradientWe 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.



Potential GradientThe general line-integral

relationship between V and E is: V d E L

V E L

dV d E LFor a very short element of length

ΔL, E is essentially constant:

cosV E L

cosdV EdL

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:

cosV EL


Potential GradientChapter 4 Energy and Potential

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.

max

dV EdL

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.



For the equipotential surfaces below, find the direction of E at P.

max

,

180

dVdL

E



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:

maxN

dVdL

E = a

max

dV dVdL dN

The maximum magnitude occurs when ΔL is in the aN direction. Thus we define dN as incremental length in aN direction,

NdVdN

E = a



Potential Gradient

Gradient of grad NdTT TdN

a

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:

Using the new term,

NdVdN

E = a grad V=



Potential GradientSince V is a function of x, y, and z, the total differential is:

But also,

V V VdV dx dy dzx y z

dV d E L x y zE dx E dy E dz

Both expression are true for any dx, dy, and dz. Thus:

xVEx

yVEy

zVEz

x y zV V Vx y z

E a a a

grad x y zV V VVx y z

a a a

Note: Gradient of a scalar is a vector.


x y zV V VVx y z

a a a


x y zx y z

a a a

V E

Introducing the vector operator for gradient:

We now can relate E and V as:

1z

V V VVz

a a a

1 1sinr

V V VVr r r

a a a

Rectangular

Cylindrical

Spherical



ExampleGiven the potential field, V = 2x2y–5z, and a point P(–4,3,6), find V, E, direction of E, D, and ρv.

22( 4) (3) 5(6)PV 66 V

V E x y zV V Vx y z

a a a 24 2 5x y zxy x a a a

24( 4)(3) 2( 4) 5P x y z E a a a 48 32 5 V mx y z a a a

0P PD E 3425 283.3 44.27 pC mx y z a a a

0div div v D E 12(8.854 10 )( 4 )y 335.42 pC my3At , 35.42(3) pC mvP 3106.26 pC m

,P

E PP

EaE


The DipoleThe 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.


The DipoleThe 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).



The DipoleThe total potential at P can be written as:

0 1 2

1 14QV

R R

2 1

0 1 24R RQR R

The plane z = 0 is the locus of points for which R1 = R2 ► The potential there is zero (as also all points at ∞).



The DipoleFor a distant point, R1 ≈ R2 ≈ r, R2–R1 ≈ dcosθ

20

cos4QdV

r

Using the gradient in spherical coordinates,V E

1 1sinr

V V Vr r r

a a a

3 30 0

cos sin2 4rQd Qd

r r

E a a

30

2cos sin4 rQdr

E a a



The DipoleTo 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.

20

cos4QdV

r

r = 2.236

r = 1.880

Plane at zero potential

45°



The DipoleThe 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.

Qp d

204rVr

p a

Since dar = d cosθ , we then have:

20

14

V

r rpr rr r

• Dipole charges: 2 3

1 1,Vr rE

• Point charge: 2

1 1,Vr rE



Exercise Problems1. 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 universityerwin sitompuleem 6/1 lecture 6 engineering electromagnetics dr.-ing. erwin...

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