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EELE 3331 – Electromagnetic I

Chapter 4

Electrostatic fields

Islamic University of Gaza

Electrical Engineering Department

Dr. Talal Skaik

2012

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Electric Potential

The Gravitational Analogy

Moving an object upward against the gravitational field increases its gravitational potential energy. An object moving downward within the gravitational field would lose gravitational potential energy.

When you move a charge in an electric field its potential energy changes. This is like moving a mass in a gravitational field.

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Electric Potential

•In order to bring two like charges near each other work must be done. In order to separate two opposite charges, work must be done. •As the monkey does work on the positive charge, he increases the energy of that charge. The closer he brings it, the more electrical potential energy it has. •The external force F against the E-field increases the potential energy.

• To move a point charge Q

a distance dl, the work done is

the product of the force and

distance :

• Negative sign indicates that work is done by an external agent.

(The force we apply to move the charge against the electric field is

equal and opposite to the force due the field).

• If we want to move the charge in the direction of the field, we do

not do the work, the field does. 4

4.7 Electric Potential

F l= QE ldW d d Displacement of point charge Q in an electrostatic field E.

The total work done in moving Q from A and B, is:

Define potential difference V as the work done (by external force) in

moving a unit positive charge from one point to another in an Electric

Field.

5

Electric Potential

W E l

B

A

Q d

finalpoint

initialpoint

AB

(Joules/coulomb)Potential W

=V = E l (J/C)Difference

(V)

B

A

d orQ

or

Notes:

• If VAB is negative, there is loss in potential energy in moving Q

from A to B→ Work is don by the field.

• If VAB is positive, there is gain in potential energy in moving Q

from A to B→ External agent performs the work.

• VAB is independent of the path taken.

6

Electric Potential

ABV E l

B

A

d

7

Electric Potential

2

0

AB 2

0

AB 2

0 0

If E is due to a point charge

Q located at the origin,

E a 4

l= a a sin a

V E l a a4

1 1V

4 4

B

A

B

A

r

r

rB

r r

A r

r

B Ar r

Q

r

d dr rd r d

Qd dr

r

Q Qdr

r r r

or

AB B A V V V

8

Electric Potential

A A

0

If we choose a zero reference for potential,

Let V=0 at infinity. Thus if V =0 as r

-4

, the potential at any point distant from a

point charge Q at the origin is:

AB B A B

Bzero

QV V V V

r

Generally r

0

0 at infinity is reference4

QV V

r

The potential at any point is the potential difference between that point

and a chosen point (or reference point) at which the potential is zero.

AB

0

1 1V

4 B A

Q

r r

0

1 2

If the point charge is located at a point with position

vector r' , V(r) becomes:

0 at infinity is reference4 '

For point charges , ,..., located at points

with position vectors

n

QV V

r r

n Q Q Q

1 2

1 2

0 1 0 2 0

1 0

, ,..., , the potential at is:

( )=4 4 4

( )=4

n

n

n

nk

k k

r r r r

Q Q QV r

r r r r r r

Qor V r

r r

9

Electric Potential

0

0

0

For continuous charge distribution, the potential

at becomes:

1 ( ') '( )= (Line Charge)

4 '

1 ( ') '( )= (Surface Charge)

4 '

1 ( ') '( )=

4 '

L

L

S

S

v

v

r

r dlV r

r r

r dSV r

r r

r dvV r

r r

(Volume Charge)

primed coordinates for source point locations. 10

Electric Potential

11

Electric Potential

The diagram shows some values of the electric potential at points in the electric field of a positively charged sphere.

→When one coulomb of charge moves from A to B it gains 15 Joules of potential energy.

12

A A

0

Notes: For E due to a point charge Q:

Assume V =0 as r Zero reference at infinity

-4

If any point other than infinity is chosen as reference

with zero potential, Let

AB B A B

Bzero

QV V V V

r

Q

0

0 0

0

/ 4 be constant=C

4 4

Generally, the potential at any point distant from a

point charge Q at the origin is: + C 4

Generall

A

AB B A B B

B Bzero

r

Q QV V V V C V C

r r

r

QV

r

y, for any E: V= E ld C

AB

0

1 1V E l

4

B

B AA

Qd

r r

13

Example 4.10

Two point charges -4μC and 5μC are located at (2,-1,3) and (0,4,-2).

Find the potential at (1,0,1), assuming zero potential at infinity.

1 2

1 20

0 1 0 2

0

1

2

6

9

Let 4 C, 5 C

V(r)= C4 4

If V( )=0, C 0

(1,0,1) (2, 1,3) ( 1,1, 2) 6

(1,0,1) (0,4, 2) (1, 4,3) 26

10 4 5 V(1,0,1)= 5.872 KV

10 6 264

36

Q Q

Q Q

r r r r

r r

r r

Hence

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Example 4.11

A point charge of 5nC is located at (-3,4,0), while line y=1, z=1 carries

uniform charge 2 nC/m.

(a) If V=0V at O(0,0,0), find V at A(5,0,1)

(b) If V=100V at B(1,2,1), find V at C(-2,5,3)

(c) If V=-5V at O(0,0,0), find VBC.

Q L

Q 12

0 0

2

0 0

0 0

Let V=V +V

V E l= a a C4 4

V E l= a a ln C 2 2

ln C2 4

r r

L LL

L

Q Qd dr

r r

d d

QV

r

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Example 4.11 - continued

0 0

0

0 0 0

0

0

9

9

ln C2 4

(a) V=0 at O(0,0,0), V at A(5,0,1)=??

1 1ln

2 4

0,0,0 0,1,1 2

5,0,1 5,1,1 1

0,0,0 3,4,0 5

5,0,1 3,4,0 9

2 10l

2 (10 / 36 )

L

LAO O A

A Aknown

A

A

AO

QV

r

QV V V

r r

r

r

V

9

9

2 5 10 1 1n

1 4 (10 / 36 ) 5 9

0 36ln 2 45(1 / 5 1 / 9) 8.477 VAO A AV V V

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Example 4.11 - continued

C B

0 0

(b) If V=100 at B(1,2,1), V at C( 2,5,3)=??

1 1V -V ln

2 4

2,5,3 2,1,1 20

1,2,1 1,1,1 1

2,5,3 3,4,0 11

1,2,1 3,4,0 21

20 1 1100 36ln 45 50.175 V

1 11 21

L C

B C B

C

B

C

B

C

Q

r r

r

r

V

V

BC

49.825 V

(C) V 49.825 100 50.175 V

We don't need a potential reference if a common reference is assumed.

C

C BV V

E l 0

E l 0 integral form

No net work is done in moving

a charge along a closed path in

an electrostatic field.

Applying Stoke's theorem

BA AB BA AB

L

L

V V V V d

d

:

E l E S=0 E=0 (conservative, irrotational)

E=0

L S

d d

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4.8 Relationship between E and V – Maxwell’s Equation

2nd Maxwell’s equation for static electric field.

(differential form)

E=0 E

0

V

V

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Relationship between E and V

Electric Field Intensity E is the gradient of V.

Negative sign shows that the direction of E is opposite to the

direction in which V increases; E is directed from higher to lower

levels of V.

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Example 4.12 Given the potential V=(10/r2)sinθ cosφ.

a) Find the electric flux density D at (2,π/2,0).

b) Calculate the work done in moving a 10µC charge from point

A(1,300,1200) to B(4,900,600).

0

3 3 3

0

0

1 1D= E, E= a a a

sin

20 10 10E= sin cos a cos cos a sin a

At (2, /2,0) D= E

20 = a 0 a 0 a

8

r

r

r

V V VV

r r r

r r r

2 2

0 =2.5 a C/m 22.1 a pC/mr r

3 3 3

' ' ' '

20 10 10E= sin cos a cos cos a sin a

(b) Method 1:

E l or E l

Break up path (to make integration easier)

E l

' a

'

r

L L

AA A B B B

r

r r r

WW Q d d

Q

Wd

Q

AA dl dr

A

' a

' sin a

B dl r d

B B dl r d

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Example 4.12 - continued (b) Calculate the work done in moving a 10µC charge from point

A(1,300,1200) to B(4,900,600).

0

00

0

4 90

3 34 30

1 30=120=120

60

34

120=90

' a , ' ' a , ' sin a

20sin cos 10cos cos

10sin + sin

75 5 10 45

32 32 16 16

r

rr

r

AA dl dr A B dl r d B B dl r d

Wdr r d

Q r r

r dr

WW

Q

6 0 0 0 0

28.125 J

Method 2:

10 1010 10 sin 90 cos60 sin 30 cos120

16 1

28.125 J

AB AB

B A

Q

WV W QV

Q

W Q V V

W

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Example 4.12 continued

0 1 2

2 1

0 1 2

2 1

2

2 1

0

The potential at point

( , , ) is:

1 1

4

4

If :

cos

co

4

P r

QV

r r

Q r r

rr

r d

r r d

Q dr r r V

2

s

r

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4.9 An Electric Dipole and Flux Lines

An electric dipole is formed when two point charges of equal

magnitude but opposite sign are separated by a small distance.

2

0

3 3

0 0

3

0

cos

4

1 1E a a a

sin

cos sinE a a

2 4

E= 2cos a sin a4

r

r

r

Q dV

r

V V VV

r r r

Qd Qd

r r

Qdor

r

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An Electric Dipole and Flux Lines

2

0

2 3

0 0

3

0

cos

4

Define dipole moment P=Qd, cos =P a

(P is directed from )

P a r P r= , a =

4 r 4

If the dipole center is at r':

P ( ')( )=

4 '

r

rr

Q dV

r

Qd

Q to Q

V Vr r

r rV r

r r

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An Electric Dipole and Flux Lines

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Equipotential Lines

• EQUIPOTENTIAL LINES are

lines along which each point is at

the same potential.

• On an equipotential surface, each

point on the surface is at the same

potential.

• The equipotential line or surface is

perpendicular to the direction of the

electric field lines at every point.

• Thus, if the electric field pattern is

known, it is possible to determine

the pattern of equipotential lines or

surfaces, and vice versa.

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Equipotential Lines In the following diagrams, the dashed lines represent equipotential lines and the solid lines the electric field lines.

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An Electric Dipole and Flux Lines

Equipotential surfaces for (a) a point charge and (b) an electric dipole.

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An Electric Dipole and Flux Lines

Notes:

• An electric flux line is an imaginary path or line and its direction

at any point is the direction of the electric field at that point.

• Equipotential surface: potential is the same at any point.

• E is always normal to the equipotential surfaces.

• Equipotential line : is intersection of equipotential surface and a

plane.

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Example 4.13

Two dipoles with dipole moments -5az nC.m and 9az nC.m are located

at points (0,0,-2) and (0,0,3). Find the potential at the origin.

1 1 1 2

3 3

0 1 2

9

1 z 1

9

2 z 2

9

3 39

1 r rV=

4

where

P 5 10 a , r (0,0,0) (0,0, 2) 2a

P 9 10 a , r (0,0,0) (0,0,3) 3a

1 10 27 = 10 20.25

2 3104

36

z

z

P P

r r

Hence V

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4.10 Energy density in electrostatic fields

Consider a region free of electric fields. Let there be three point

charges Q1,Q2,Q3 at infinity

To determine the energy present in the assembly of charges, we

have to determine the amount of work necessary to assemble

them.

No work is required to transfer Q1 from infinity to P1 because the

space is initially charge free (No electric field)

The work done in transferring Q2 from infinity to P2 is equal to the

product of Q2 and the potential V21 at P2 due to Q1.

The work done in positioning Q3 at P3 is equal to Q3(V32+V31).

The total work in positioning the three charges is:

V21→Potential at point 2 due to Q1.

V31→Potential at point 3 due to Q1.

V32→Potential at point 3 due to Q2.

If charges were positioned in reverse order:

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Energy density in electrostatic fields

1 2 3

2 21 3 31 320 (1)

EW W W W

Q V Q V V

3 2 1

2 23 1 12 130 (2)

EW W W W

Q V Q V V

Adding equations (1) and (2)

V1, V2, V3→Total potentials at points 1,2,3 respectively.

If there are n point charges:

In case of continuous charge distribution, summation becomes

integration:

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Energy density in electrostatic fields

1 12 13 2 21 23 3 31 32

1 1 2 2 3 3 1 1 2 2 3 3

2

12 = + +

2

E

E E

W Q V V Q V V Q V V

W Q V Q V Q V or W QV Q V Q V

1

1

2

n

E k k

k

W Q V

1 (Line Charge)

2

1 (Surface Charge)

2

1 (Volume Charge)

2

E L

L

E S

S

E v

v

W Vdl

W VdS

W Vdv

Define the electrostatic energy density wE (in J/m3)

(Energy Stored)

33

Energy density in electrostatic fields

22

0

0

2

0

1 1D E =

2 2 2

,

1 1D E

2 2

EE

E E

v

E

v v

dW Dw E

dV

so W w dv

W dv E dv

34

Example 4.14

The point charges -1nC, 4nC, 3nC are located at (0,0,0), (0,0,1), and

(1,0,0). Find the energy in the system.

35

Example 4.15

(section 4.6 D)

(The energy stored)

36

Example 4.15

37