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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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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.
V(r) = 14oV
r.P(r)r2 d
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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.
V(r) = 14oV
r.P(r)r2 d
(1/r) =r/r2 ( w.r.t. source coordinate)
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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.
V(r) = 14oV
r.P(r)r2 d
(1/r) =r/r2 ( w.r.t. source coordinate)
V(r) = 14o VP.(1/r)d
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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.
V(r) = 14oV
r.P(r)r2 d
(1/r) =r/r2 ( w.r.t. source coordinate)
V(r) = 14o VP.(1/r)d
Use the product rule: .(fA) = f(.A) + A.(f)
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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.
V(r) = 14oV
r.P(r)r2 d
(1/r) =r/r2 ( w.r.t. source coordinate)
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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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.
V(r) = 14oV
r.P(r)r2 d
(1/r) =r/r2 ( w.r.t. source coordinate)
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
Divergence Theorem:V
(.v) d=Sv.da
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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.
V(r) = 14oV
r.P(r)r2 d
(1/r) =r/r2 ( w.r.t. source coordinate)
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
Divergence Theorem:V
(.v) d=Sv.da
V(r) = 14o S1rP.da 14o V
1r(.P) d
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Field of a polarized object
V(r) = 14o
S
1r
P.da 14o
V
1r
.P
d
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Field of a polarized object
V(r) = 14o
S
1r
P.da 14o
V
1r
.P
d
V(r) = 14o b
rda + 14o
brd
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Field of a polarized object
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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Field of a polarized object
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.
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
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Field of a polarized object
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.
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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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.
f: Free charge density within the dielectric.=f +b
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Th El t i Di l t
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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.
f: Free charge density within the dielectric.=f +b
Gausss law: o.E== .P +f . (oE + P) =f
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Th El t i Di l t
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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.
f: Free charge density within the dielectric.=f +b
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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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.
f: Free charge density within the dielectric.=f +b
Gausss law: o.E== .P +f . (oE + P) =f Define electric displacement DasD= oE + P
.D= f In integral form:D.da= Qfenc
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The Electric Displacement
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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.
f: Free charge density within the dielectric.=f +b
Gausss law: o.E== .P +f . (oE + P) =f Define electric displacement DasD= oE + P
.D= f In integral form:D.da= Qfenc
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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The Electric Displacement
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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.
f: Free charge density within the dielectric.=f +b
Gausss law: o.E== .P +f . (oE + P) =f Define electric displacement DasD= oE + P
.D= f In integral form:D.da= Qfenc
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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The Electric Displacement
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The Electric Displacement
For a linear dielectric, D= oE + P=o(1 +e)E= E
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The Electric Displacement
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The Electric Displacement
For a linear dielectric, D= oE + P=o(1 +e)E= E
=o(1 +e) =or
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The Electric Displacement
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The Electric Displacement
For a linear dielectric, D= oE + P=o(1 +e)E= E
=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
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.
b= .P= z(az
2 +b) = 2az
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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.
b= .P= z(az
2 +b) = 2az
b |z=L=P.z=aL2 +b
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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.
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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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.
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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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.
b= .P= 1r2
rr
2(kr)
= 1r2
(3kr2) = 3k
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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.
b= .P= 1r2
rr
2(kr)
= 1r2
(3kr2) = 3k
Total bound volume charge= (3k)43R2 = 4kR3
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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.
b= .P= 1r2
rr
2(kr)
= 1r2
(3kr2) = 3k
Total bound volume charge= (3k)43R2 = 4kR3
b=P.n= kr.r= kRr.r=kR
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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.
b= .P= 1r2
rr
2(kr)
= 1r2
(3kr2) = 3k
Total bound volume charge= (3k)43R2 = 4kR3
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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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.
b= .P= 1r2
rr
2(kr)
= 1r2
(3kr2) = 3k
Total bound volume charge= (3k)43R2 = 4kR3
b=P.n= kr.r= kRr.r=kR Total bound surface charge = kR(4R2) = 4kR3
For r< R, draw a spherical Gaussian surface at radiusr< R.
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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.
b= .P= 1r2
rr
2(kr)
= 1r2
(3kr2) = 3k
Total bound volume charge= (3k)43R2 = 4kR3
b=P.n= kr.r= kRr.r=kR Total bound surface charge = kR(4R2) = 4kR3
For r< R, draw a spherical Gaussian surface at radiusr< R.
For this,SE.da= E.4r2 = 1oQenc. = 1o(43r3b)
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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.
b= .P= 1r2
rr
2(kr)
= 1r2
(3kr2) = 3k
Total bound volume charge= (3k)43R2 = 4kR3
b=P.n= kr.r= kRr.r=kR Total bound surface charge = kR(4R2) = 4kR3
For r< R, draw a spherical Gaussian surface at radiusr< R.
For this,SE.da= E.4r2 = 1oQenc. = 1o(43r3b)
E= 13o rbr= k
or (Where, b= 3k)
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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.
b= .P= 1r2
rr
2(kr)
= 1r2
(3kr2) = 3k
Total bound volume charge= (3k)43R2 = 4kR3
b=P.n= kr.r= kRr.r=kR Total bound surface charge = kR(4R2) = 4kR3
For r< R, draw a spherical Gaussian surface at radiusr< R.
For this,SE.da= E.4r2 = 1oQenc. = 1o(43r3b)
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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A dielectric cube of side a, centered at the origin, carries apolarization P= kr, where k is a constant. Find all the boundcharges.
b= .P= k.r= 3k
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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.
b= .P= k.r= 3k
Total volume charge = 3ka3
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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.
b= .P= k.r= 3k
Total volume charge = 3ka3
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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A dielectric cube of side a, centered at the origin, carries apolarization P= kr, where k is a constant. Find all the boundcharges.
b= .P= k.r= 3k
Total volume charge = 3ka3
b |z=a/2=P.z |z=a/2= kz |z=a/2=ka/2;
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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A dielectric cube of side a, centered at the origin, carries apolarization P= kr, where k is a constant. Find all the boundcharges.
b= .P= k.r= 3k
Total volume charge = 3ka3
b |z=a/2=P.z |z=a/2= kz |z=a/2=ka/2;
b |z=a/2=P. z |z=a/2= kz |z=a/2= ka/2;
Total surface charge = 3ka a2
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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.
b= .P= k.r= 3k
Total volume charge = 3ka3
b |z=a/2=P.z |z=a/2= kz |z=a/2=ka/2;
b |z=a/2=P. z |z=a/2= kz |z=a/2= ka/2;
Total surface charge = 3ka a2
Total bound charges= 0.
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