E-field of a polarized object 1

Electric field of a Polarized Object

© Frits F.M. de Mul

E-field of a polarized object 2

Electric field of a Polarized Object

Question:

Calculate E-field produced BY (not: IN) the polarized object.

A dielectric object will become polarized.

Available:

An external E-field: Eex. Eex

E-field of a polarized object 3

Electric field of a Polarized Object

• Analysis and symmetry

• Approach to solution

• Calculations

• Conclusions

E-field of a polarized object 4

Analysis and SymmetryZ

YX

Coordinate axes:

assume Z-axis // EexEex

Assume: p // Eex ;

n and p homogeneous

Result of polarization:

Dipole distribution:

n dipoles/m3 ;

each dipole moment p [Cm]

E-field of a polarized object 5

Approach to solution

Approach:

= calculate potential V ;

= E from V by differentiation

Eex

Z

YX

v

Question: calculate E-field in arbitrary point P outside v

P

Distributed dipoles:

dV- integration over volume elements dv filled with dp.

E-field of a polarized object 6

Calculations (1)

Eex

Z

YX

P

204

cos

r

dpdV

πεθ=

Z'

rer

θ

dp

dz

volume element dv with dp = np.dv = np.dS⊥..dz ; with dS⊥ ⊥ Z-axis

204

cos..

r

dzdSnp

πεθ= ⊥

204 r

Vπε•= rep

Dipole:

E-field of a polarized object 7

Calculations (2)

Eex

Z

Y

P

θπε

⊥ cos4

..2

0r

dzdSnpdV

Z'

X

rer

θ

dp

dz

Pr

dS⊥

θ

cross section through P and Z’-axis :

rb

rt

zt : top

zb : bottom

E-field of a polarized object 8

Calculations (3)

θπε

= ⊥ cos4

..2

0r

dzdSnpdV

P

rb

rt

r

zt

zb

dS⊥

θ

dz-integration ⇒ dr-integration :

dr = -dz.cosθ ⇒ dz = -dr / cosθ

204

..

r

drdSnpdV

πε−= ⊥

r

θdz

drP

E-field of a polarized object 9

Calculations (4)

P

rb

rt

r

zt

zb

dS⊥

θ

r

θdz

dr

204

..

r

drdSnpdV

πε−= ⊥

∫∫ ∫ πε−= ⊥

S

r

r

t

br

drnpdSV

204

.

P

“Polarization” P = np

∫∫

−

πε= ⊥

S bt rr

PdSV

11

4 0

E-field of a polarized object 10

Calculations (5)

∫∫

−

πε= ⊥

S bt rr

PdSV

11

4 0Eex

Z

Y

P

Z'

X

rer

θ

dp

dz

P dS P.dS⊥ = P.dS

bound charges: dσb = P.dS

∫∫

−

πεσ=

S bt

b

rr

dV

11

4 0

E-field of a polarized object 11

ConclusionsConclusions

the end

bound charges: dσb = P.dS

∫∫

−

πεσ=

S bt

b

rr

dV

11

4 0

P

Eex

Z

YX

rbottom

rtop

+

-Conclusion: the field of a polarized volume is equivalent to that of bound surface charges (provided homogeneous polarization)

+

-

+

--

-

+ ++

+ + ++

-- - -