Electrostatics Media 1

8/13/2019 Electrostatics Media 1

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Electric Fields in Matter

January 24, 2014

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Polarization

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Polarization

When a dielectric is placed in an external electric field, it

becomes polarized.

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Polarization

When a dielectric is placed in an external electric field, itbecomes polarized.

The resulting dipole moment per unit volume is referred to asPolarization P.

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Polarization

When a dielectric is placed in an external electric field, itbecomes polarized.

The resulting dipole moment per unit volume is referred to asPolarization P.

Original field which is responsible for P.Field which is due to P.

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Field of a polarized object

The potential at point X due tosingle dipole p:V(r) = 14o

p.rr2

p=Pd;P = dipole moment per unitvolume.

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p.rr2


V(r) = 14oV

r.P(r)r2 d

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p.rr2


V(r) = 14oV

r.P(r)r2 d

(1/r) =r/r2 ( w.r.t. source coordinate)

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p.rr2


V(r) = 14oV

r.P(r)r2 d


V(r) = 14o VP.(1/r)d

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p.rr2


V(r) = 14oV

r.P(r)r2 d


V(r) = 14o VP.(1/r)d

Use the product rule: .(fA) = f(.A) + A.(f)

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p.rr2


V(r) = 14oV

r.P(r)r2 d


V(r) = 14o VP.(1/r)d

Use the product rule: .(fA) = f(.A) + A.(f) V(r) = 14o

V.

Pr

d

V

1r(.P)

d

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p.rr2


V(r) = 14oV

r.P(r)r2 d


V(r) = 14o VP.(1/r)d


V.

Pr

d

V

1r(.P)

d

Divergence Theorem:V

(.v) d=Sv.da

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http://find/


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p.rr2


V(r) = 14oV

r.P(r)r2 d


V(r) = 14o VP.(1/r)d


V.

Pr

d

V

1r(.P)

d

Divergence Theorem:V

(.v) d=Sv.da

V(r) = 14o S1rP.da 14o V

1r(.P) d

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V(r) = 14o

S

1r

P.da 14o

V

1r

.P

d

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V(r) = 14o

S

1r

P.da 14o

V

1r

.P

d

V(r) = 14o b

rda + 14o

brd

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V(r) = 14o

S

1r

P.da 14o

V

1r

.P

d

V(r) = 14o b

rda + 14o

brd

b=P.n; b= .P (b and bare not fictitious.

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V(r) = 14o

S

1r

P.da 14o

V

1r

.P

d

V(r) = 14o b

rda + 14o

brd


Thus the potential of a polarized object is the same as thatproduced by a volume charge density band surface chargedensityb. (Bound charges in contrast to free charges in aconductor)

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f
http://find/


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V(r) = 14o

S

1r

P.da 14o

V

1r

.P

d

V(r) = 14o b

rda + 14o

brd


Thus the potential of a polarized object is the same as thatproduced by a volume charge density band surface chargedensityb. (Bound charges in contrast to free charges in aconductor)

Vbd

=

(.P) d

; Apply Divergence theorem:=

P.nda =

bda

Conservation of charge.

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Th El i Di l


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The Electric Displacement

In the presence of dielectrics, we have

(a) Field due to bound charges: b= .P(b) Field due to everything else, which we call free charges.

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Th El i Di l


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f: Free charge density within the dielectric.=f +b

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Th El t i Di l t


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Gausss law: o.E== .P +f . (oE + P) =f

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Th El t i Di l t
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Gausss law: o.E== .P +f . (oE + P) =f Define electric displacement DasD= oE + P

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The Elect ic Dis lace e t
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.D= f In integral form:D.da= Qfenc

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It is the Gausss law in the presence of a dielectric. It can beused to determine D from free charges if symmetry is present.

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It is the Gausss law in the presence of a dielectric. It can beused to determine D from free charges if symmetry is present.

For a linear dielectric P= oeEeelectric susceptibility of the dielectric.

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For a linear dielectric, D= oE + P=o(1 +e)E= E

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=o(1 +e) =or

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=o(1 +e) =or

: Permittivity of the mediumoPermittivity of free spacer: Relative permittivity or dielectric constant

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Problem
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Problem

A thin rod of cross section A extends along the z-axis fromz= 0 to z= L. The polarization of the rod is given byP= (az2 +b)z. Find b and b.

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Problem
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Problem


b= .P= z(az

2 +b) = 2az

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Problem
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Problem


b= .P= z(az

2 +b) = 2az

b |z=L=P.z=aL2 +b

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Problem
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Problem


b= .P= z(az

2 +b) = 2az

b |z=L=P.z=aL2 +b b |z=0=P. z= b

b |curved=P.s= 0

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Problem
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b= .P= z(az

2 +b) = 2az

b |z=L=P.z=aL2 +b b |z=0=P. z= b

b |curved=P.s= 0

Total charge =A L

0 bdz+ bda=A[2az2

2 |L0

] +A[(aL2 +b) b] = A(aL2) +AaL2 = 0

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Problem 4.10
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A sphere of radius Rcarries a polarization P(r) =kr where kis a constant and r is the vector from the center.

(a) Calculate the bound charges b and b.(b) Find the field inside and outside the sphere.

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Problem 4.10
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b= .P= 1r2

rr

2(kr)

= 1r2

(3kr2) = 3k

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Problem 4.10
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b= .P= 1r2

rr

2(kr)

= 1r2

(3kr2) = 3k

Total bound volume charge= (3k)43R2 = 4kR3

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Problem 4.10
http://find/


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b= .P= 1r2

rr

2(kr)

= 1r2

(3kr2) = 3k


b=P.n= kr.r= kRr.r=kR

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Problem 4.10
http://find/


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b= .P= 1r2

rr

2(kr)

= 1r2

(3kr2) = 3k


b=P.n= kr.r= kRr.r=kR Total bound surface charge = kR(4R2) = 4kR3

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Problem 4.10
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b= .P= 1r2

rr

2(kr)

= 1r2

(3kr2) = 3k



For r< R, draw a spherical Gaussian surface at radiusr< R.

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Problem 4.10
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b= .P= 1r2

rr

2(kr)

= 1r2

(3kr2) = 3k




For this,SE.da= E.4r2 = 1oQenc. = 1o(43r3b)

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Problem 4.10
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b= .P= 1r2

rr

2(kr)

= 1r2

(3kr2) = 3k





E= 13o rbr= k

or (Where, b= 3k)

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Problem 4.10
http://find/


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b= .P= 1r2

rr

2(kr)

= 1r2

(3kr2) = 3k





E= 13o rbr= k

or (Where, b= 3k)

For r> R, same as if all charge at center; but Qtot= 0, soE= 0

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Problem 4.31
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A dielectric cube of side a, centered at the origin, carries apolarization P= kr, where k is a constant. Find all the boundcharges.

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Problem 4.31
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b= .P= k.r= 3k

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Problem 4.31
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b= .P= k.r= 3k

Total volume charge = 3ka3

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Problem 4.31
http://find/


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b= .P= k.r= 3k


b |z=a/2=P.z |z=a/2= kz |z=a/2=ka/2;

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Problem 4.31
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Electrostatics Media 1