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Chapter 4. Potential and energy

1

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2Solving procedure for EM problems

Known charge distribu-tion

Coulomb’s law

Known boundary condi-tion

Gauss’ law differential form

/DED

Vector cal-culation

Vector cal-culation

V R

d2

04

ˆ

R

E

Known charge distribu-tion

Integration of Coulomb’s law

Scalar cal-culation

V R

dV

04

Known charge distribu-tion

Poisson equationScalar cal-culation

0

2

V

VE

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Gravitational field

rF ˆ2r

GMm

rF

G ˆ2r

GM

m

Earth

Moon

3

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4Gravitational potential

Instead of field lines, potential energy levels can imply the direction and magnitude of gravitational force.

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2r

2r

Work in a gravitational field

2

2

)(r

rrF dW

122rF

r

GMm

1r

To move an object in the gravitational field, an external force must be ap-plied that compensates the force due to gravity.

O

22

2222

22

2

.)(

111

ˆˆ

2

2

2

2

r

GMgrmgrr

r

GMm

rrGMm

rGMm

drr

GMmW

r

r

r

r

rrFMmass :

mmass :

5

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Potential energy in a gravitational field

• The scalar field quantity of potential energy is intro-duced to represent energy levels inherent to positions in the space.

• Differences between the energy levels can be ob-tained from the work that must applied for the object to move from the initial position to the final position.

• The position that corresponds to zero energy level is one that is located far away from the earth.

)( 2rU

)( 2rU

2r

2r

1r

O

WUUU )()( 22 rr

6

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7Potential energy in an electric field

Electric fieldElectric potential Electric potential (3D)

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Electric potential energy (V)

+q

VqdqdW tt

B

A

B

A

r

r

r

rrErF )()(

+qt

• As in the gravitational field, a potential energy for the electric field can be introduced.

• The potential energy in an electric field is defined as the energy levels of the test charge with +1C.

• The unit of potential energy is “voltage” named after the physicist Volta.

• The position far away from the source charge has zero potential energy.

Electric potential energy defined as the work to move a test charge with (+1C)

B

A

dVr

rrE)(

8

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Electric potential due to a point charge

ABAB

r

r

r

r

B

A

AB VVr

q

r

q

r

qdr

r

qdV

B

A

B

A

0

1

0

1

0

12

0

1

444ˆˆ

4 rrsE

+q1

ArBr r

r

q

r

qdr

r

qdV

rrB

A

AB0

1

0

12

0

1

44ˆˆ

4

rrsE

9

If the position A is infinitely distant from the charge q1, VA approaches zero.

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+q1

+q1

r

qdV

r

0

1

4)(

rE

Distribution of electric potentials due to a point charge 10

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+q -q

Potential distribution due to a dipole

R

q

R

qV

00 44

11

rrrr RR ,

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A positive charge moves from the position of high potential energy to that of lower potential energy.

+q1

Movement of a charge in a potential field

+q1

12

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13Structure of a cathode ray tube

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B

A

dV rE

rd

Scalar field V

dzzdyydxx

zzyyxxdd

ˆˆˆ

)ˆˆˆ(

r

dzzdd

dzzdd

zzdd

ˆˆˆ

ˆˆˆ

)ˆˆ(

r

drdrdrr

dd

rdd

d

rdrdrr

rdrdrrrrdd

sinˆˆˆ

ˆˆˆ

ˆˆ)ˆ(

r

• Rectangular coordinate

• Cylindrical coordinate

• Spherical coordinate

• To obtain the potential difference V, we should in-tegrate the electric field from A to B.

AB

B

A

VVdV rE

AAV

BBV

• For every position, the potential V is defined. So the potential is a scalar field quantity.

• Only the potential difference has physical signifi-cance. So voltage reference point should be speci-fied always.

14

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Example 4.1

Vdyydxxdzdyxdxy

dzdydxxydV

xy

B

A

B

A

B

A

AB

48.0112

ˆˆˆˆ2ˆˆ

ˆ2ˆˆ

6.0

0

28.0

1

2

zyxzyxsE

zyxE

AAV

BV

1

1

x

y

)1,0,1(A

)1,6.0,8.0(B

AAV

BV

1

1

x

y

)1,0,1(A

)1,6.0,8.0(B

Vdyy

dxx

dzdyxdxydV

xyB

A

B

A

AB

48.03

1)1(3

2

ˆ2ˆˆ

6.0

0

8.0

1

sE

zyxE

sd

sd• In this example, a different integration path is used with the

same start point A and end point B.

• Although the integration paths are different, the voltage differ-ence is the same. A field that has this property is called as a con-servative field.

15

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Conservative field

21 CC

AB ddV sEsE

AAV

BBV

1C

2C

3C

• All electric field in electrostatic problems are conservative field.

• In conservative field, the voltage difference de-pends on only the start and the end point. The re-sult of the integral is independent of the path.

• The condition for a conservative field is that curl of the field should be zero.

fieldveconservati0

laws'Faraday;t

이면

E

BE

AAV

BBV

1C

2C

21

21

21

0

00

CC

CC

CC C

C

dd

dd

ddd

d

sEsE

sEsE

sEsEsE

sEEE

2C

y

E

x

Eˆ

x

E

z

Eˆ

z

E

y

Eˆ xyzxyz zyxE

• The electric field in the example 4.1 is conservative.

16

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17

Conservative fieldNon-conservative field

Circulating elec-tric field is non-conservative.

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Relation between E and V

C

AB dV sE

AAV

BBV

1C

• We have learned how to obtain the voltage difference from an electric field. The re-verse process is also possible. That is, we can obtain the electric field from the poten-tial distribution.

• As in the derivation process for divergence operator, the relation between the electric field and the potential can be derived from the integral equation with the integration path infinitesimally small.

• Compared with electric field calculations which contain vector operations, voltage calculations are easier and simpler as the voltage is scalar quantity.

• For ease of operation, we calculate first potential functions and then electric field can be derived from potentials.

sd

AV

BV

VE

1

,,

),,(

331

hhh

z

VE

y

VE

x

VE

dzz

Vdy

y

Vdx

x

VzyxdV

dzEdyEdxEddVVV

zyx

zyxAB sE

• This operation is called “gradient V”.

dV

18

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ii

iii

i

ii

iiiii

iiiii

iiAB

u

V

hV

u

V

hE

duu

VduhEdsE

duhdsduu

VdV

dsEddVVV

1,

1

,

sE

• Gradient operators in other coordinate systems can be derived from the following relation.

1,,1

ˆˆ1

ˆ

,1

,

321

hhh

z

VVVV

z

VE

VE

VE

dzz

Vd

Vd

VdV

dzEdEdEddVVV

z

zAB

zφρ

sE

sin,,1

ˆsin

1ˆ1ˆ

sin

1,

1,

sin

321 rhrhh

V

r

V

rr

VV

V

rE

V

rE

r

VE

dV

dV

drr

VdV

drEdrEdrEddVVV rAB

φθr

sE

Cylindrical coordinate Spherical coordinate

Gradient operator in other curvilinear coordinates

rE ddV

19

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Properties of gradient operator

+q1 -q1

Equi-potential surface

. lar toperpendicu is

,zero isproduct inner theBecause

0

(constant))(

r

r

r

dV

dVdzz

Vdy

y

Vdx

x

V

CV

sd

EV

• Because the derivatives of voltage on the equi-potential surfaces are zero, Gradient V is perpendicu-lar to those surfaces.

• Electric field line is directed from the higher potential region to lower region.

• Gradient V is directed to higher potential region.

V5V3

RdR

dfy

RRfy

ˆ

'),(

rr

V0

V5

Using chain rule, gradient operation becomes simpler. For the function f of argument R

20

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Potential of multiple charge distribution

+q1

+q2

+q3n0

n

20

2

10

1

i i0

iii

R

i2

i0

iR

n2

n0

n22

20

212

10

1

4

q

4

q

4

qV

r4q

drˆˆr4

qdV

4

q

4

q

4

q)(

rrrrrr

rrsE

arr

arr

arr

rE

+qn

)(V r

Continuous charge distribution

'dr4

)(V

'dr4

)('dd

'4

ˆ)(d'd

'4

ˆ)(dV

'd'4

ˆ)()(

'V 0

'V 0'V C2

0C 'V2

0C

'V2

0

r'

r's

rr

Rr's

rr

Rr'sE

rr

Rr'rE

)(r

)(V r

'dr4

)(V

'C 0

L l

r'

• If potential contributions of separate charges are added, the poten-tial of the multiple charges are obtained.

• If the charge distribution is continuous, the sum (Σ) symbol is replaced with integral (∫) symbol.

'dar4

)(V

'S 0

S

r'• Line charge: • Surface

charge:

Point charges

21

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22

Sb

a

z

ρrzrrrR ˆ,ˆ, z

2222

00

2/1220

2

0 2/1220

' 0

22

'

2

'4

''4

)(),0(

22

22azbzdt

z

d

ddz

dazV

Sbz

az

S

b

a

S

b

a

S

S

S

rr

r'

Example : potential due to a charged annular disk

Find voltage on the z-axis, and electric field using the voltage.

22 zt

22220

11

2ˆ),0(

bzaz

zVz S

zE

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Electric dipole

+q

-q

)cosr,sinr,0( P

)2/d,0,0(Q

)2/d,0,0(Q

sinˆcos2ˆr4

cosqdV

r4

ˆ

r4

cosqd

r

PQPQ

4

q

PQPQ

PQPQ

4

q

PQ

1

PQ

1

4

q

PQ4

q

PQ4

qV

30

20

20

20

0000

θrE

rp

cosdcosr

d1rcos

r

d1rcosrdrcosrdr

)2/d(cosrdr)2/d(cosrdrPQPQ

)2/d(cosrdr)2/dcosr()sinr()2/dcosr,sinr,0(PQ

)2/d(cosrdr)2/dcosr()sinr()2/dcosr,sinr,0(PQ

22

2222

2222

2222

)d,0,0(d

d

.momentdipole;qdp

(Equi-potential surface)

θ

z

23

xx2

111

300

200000 ))((

!3

1))((

!2

1))(()()( xxxfxxxfxxxfxfxf

EMLABEMLAB

24Poisson’s & Laplace’s equations

EDD ,

)equationsPoisson'(2

V

VE

)()( VE

VV 2

for homogeneous medium

2

2

2

2

2

2

ˆˆˆˆˆˆ

ˆˆˆ

z

V

y

V

x

V

z

V

y

V

x

V

zyx

z

V

y

V

x

VV

zyxzyx

zyx

Laplace operator has different forms for different coordinate systems.

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25

The differential equation for source-free region becomes a Laplace equation.

)equations'Lapace(0V2

2

2

2

2

22 11

z

VVVV

2

2

2

2

2

22

z

V

y

V

x

VV

2

2

2222

22

sin

1sin

sin

11

V

r

V

rr

Vr

rrV

(rectangular coordinate)

(cylindrical coordinate)

(spherical coordinate)

Laplace’s equations

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26Example 1 : Laplace eqs.

Sd

zx

0V

V0

Unlike the procedures in the previous chapters, the potential V is first obtained solving Laplace equation. Then, using the potential, E, D, , Q, C are obtained.

02

2

2

2

2

2

2

22

zzyx

d

S

dV

SV

V

QC

d

SVSdaQ

d

Vd

Vd

Vd

V

zd

Vz

s

S

s

s

s

0

0

0

0

0

0

0

0

)6(

)5(

ˆˆbottom)(

ˆˆtop)()4(

ˆ)3(

ˆ)2(

)()1(

DzDn

DzDn

zED

zE

If the plates are wide enough to ignore the variation of electric field along x and y directions

0,0

yx

constant),( BABAzAz

zd

Vz

d

VA

BVBAdd

00

0

)(

0)0(,)(

Using the boundary conditions on the two plates,

EMLABEMLAB

27Example 2

)/ln(

)/ln()(

)/ln(

ln,

)/ln(

ln)(,0ln)(

,conditionsboundary theUsing

constants),(ln,0

0,0

symmetricaxially anddirectionz in the infinite be toassumed iscylinder The

0111

000

0

2

2

2

2

2

2

2

2

22

ab

bVV

ba

bVB

ba

VA

VBaAaVBbAbV

BABAVAVV

z

VV

V

z

VVVV

)/ln(

2)6(

2)/ln(

1)5(

)/ln(

1ˆˆ)(

)/ln(

1ˆˆ)()4(

ˆ)/ln(

1)3(

ˆ)/ln(

1)2(

)/ln(

)/ln()()1(

0

0

0

0

0

0

ab

L

V

QC

aLaba

VdaQ

abb

Vb

aba

Va

ab

V

ab

VV

ab

bVV

S

s

s

s

DρDn

DρDn

ρED

ρE

x

y

r

a

b

V00V

EMLABEMLAB

28Charge storage

Q V

+V-

If charges are accumulated, potential difference increases.

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29

V

QC

h

VolumeS

Capacitor

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30

0V

0Q

Due to potential difference, positive charges rush to the capacitor. As the amount of charges increases, the voltage increases.

If the voltage difference between the terminals of the capacitor is equal to the supply voltage, net flow of charges becomes zero.

Charging capacitor

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31

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32Potential distribution near parallel plates

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Electrostatic energy

+q1

+q2

+q3

N

n

n

iinn

N

nntotal

n

iinn

n

iinnn

VqWW

VqVqW

VqVqVqW

VqVqW

VqW

2

1

1,

2

1

1,

1

1,

3,442,441,444

2,331,333

1,222

1

1 1,

1

1

1,

1,

,111

,224,223,222

,114,113,112,111

N

n

N

niinn

N

nntotal

N

niinn

N

niinnn

NNNN

N

N

VqWW

VqVqW

VqW

VqVqVqW

VqVqVqVqW

N

nnntotal

N

n

N

nii

innnn

N

n

N

nii

inn

N

n

N

niinn

n

iinn

N

n

N

niinn

N

n

n

iinntotal

VqW

VVVqVq

VqVq

VqVqW

1

1 1,

1 1,

1 1,

1

1,

1 1,

1

1

1,

2

1

2

N

nnnVqW

12

1

• The work to assem-ble charges q1, q2~

qn.

• The work for charges q1, q2,~,qn to be sepa-

rated to infinitely dis-tant points.

• Vn is a potential due to N-1

charges other than n-th charge

The magnitudes of works to assemble or disassem-ble are the same.

33

ii

jji

qV

rr

0, 4

Potential energy of qi due to qj.

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Electrostatic energy

')(2

1'

2

1')(

2

1

')(2

1')(

2

1

')'()'(2

1

''

''

'

ddVdV

dVVdV

dVW

VVS

VV

V

S

DEDaD

DDD

rr

AAA uu)u(

'd)(2

1W

'V

DE

• If the product V*D becomes zero on the surface S, the surface integral becomes zero.

0

Vq

ba

qdr

r

rqdrr

r

q

ddrdrdW

r

q

ba

qaqVW

br

q

r

drrrqrV

b

a

b

a

b

a

b

a

r

b

11

8832

4

sin2

1'

2

1

,ˆ4

)2(

11

8)(

2

1

11

44

ˆˆ)()1(

2

4

222

4

2

2

2

0 0

2

2

0

2

02

0

DDDE

rD

0

Vq

(1) The integral is performed on the surfaces marked by red lines.

(2) The integral is performed over the vol-ume marked by blue lines.

34

AA

A

uu

z

AuA

z

u

y

AuA

y

u

x

AuA

x

u

z

uA

y

uA

x

uAu

zz

yy

xx

zyx )()()()(

EMLABEMLAB

35Capacitance• The magnitude of an electric field is proportional to charges, and voltages are propor-

tional to electric field. Hence, charges are proportional to voltages. This proportionality constant is called capacitance.

CVQVQQEV

VV

d

d

d

V

QC SS

aE

sE

aE

SS

S

d

d

SC

d

S

dz

da

dz

da

d

d

V

QC d

S

S

S

dS

S

S

S

S

00

1ˆˆ

)ˆ(ˆ

ˆ

zz

zz

sE

aE

zEz

x

Example: Capacitance of a parallel plate capacitor

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36Capacitance from electrostatic energy

22

22

2

2

1

2

1

2

1

2

1

2

1)2(

2

1

2

1)1(

sE

EE

EEED

d

d

V

WC

CVQC

QCVQVdVdVW

ddW

Ve

VV

e

VV

e

SS

S

dzx

d

SC

d

S

d

Sd

V

W2C

ddzˆˆV

Sd2

d2

1d

2

1W

ˆ

2

S

2S

2e

Sd

0

S

2S

V

2

S

V

e

S

zz

EE

zE

Example : parallel plate capacitor