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Electromagnetism Physics 15b

Lecture #22 Maxwells Equations in Dielectrics

Purcell 10.1110.15

What We Did Last Time A dipole generates electric field A dipole in an electric field receives: Torque If E is non-uniform, net force Dipoles are attracted to stronger E field

Density of polarization P = Np Small volume dv of dielectric looks like a dipole Pdv A cylinder parallel to P looks like charge density P

on the ends Average electric field inside the cylinder is E = 4P

A uniformly polarized sphere External field looks like produced by a dipole VP Internal field E = (4/3)P

N = p E F = (p )E

Er =

2p cosr 3

,E =p sin

r 3

2

Todays Goals Connect polarization P and the dielectric constant Continue discussion of polarized sphere Boundary condition at the surface How to make a sphere polarized uniformly

Examine the effects of non-uniform polarization Non-uniform P creates bound charge distribution Charge screening, electric displacement

Discuss time-dependent E field in dielectrics Modified Maxwells equations Electromagnetic waves in dielectrics

Dielectric Constant How does P relate to the dielectric constant ? Consider the filled capacitor again Electric field is reduced by factor 1/

Polarization density P is related to the average electric field E that causes it by: This is an empirical law, and

is correct within limits

Q +Q

d

E =

4

= 4 4P

Field from charge on the plates

Field from polarization

E = E 4P PE

= 14

eelectric susceptibility

P = eE =

14

E

3

Uniformly Polarized Sphere A dielectric sphere is uniformly polarized along +z It contains dipoles p = qs with density N

This can be seen as two overlapping spheres Charge densities are +Nq and Nq Centers are separated by distance s

From outside, each sphere looks like a point charge (recall Gauss) +NqV and NqV

Field outside is identical to that generated by a single dipole moment NqVs = V P We know this field

from the last lecture:

P = Np = Nqs

(r > R) = V P r

r 2=

VP cosr 2

R

Uniformly Polarized Sphere Inside the sphere, there is no net charge The field must obey Laplaces eqn. We also need the boundary condition,

i.e., values of at the surface, which we know from the outside field

A uniform electric field along +z works

The field due to a uniformly polarized sphere is Inside: E = (4/3)P Outside: identical to the field generated by a dipole (4/3)R3P

(r = R) = VP cos

R2=

VR3

Pr cos = 43

Pz

(r < R) = 4

3Pz

E(r < R) = 4

3P

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Boundary Conditions Electric potential is a continuous function of space Otherwise there would be infinite electric field

As a result, electric field has to satisfy the following conditions on the surface of the dielectric E|| parallel to the surface is continuous E perpendicular to the surface

may be discontinuous Check this with the uniformly polarized sphere:

Er =83

P cos

E =43

P sin

Outside

Er = 43

P cos

E =43

P sin

Inside

Dielectric Sphere in E Field How does a dielectric sphere get uniformly polarized? Try the simplest way put it in a uniform external field E0 Suppose this leads to a uniform polarization P P generates a uniform field inside:

Total field inside is Resulting polarization is

We can solve this to find

Uniform external field polarizes the sphere uniformly

E = 4

3P

Einside = E0 + E = E0

43

P

P = eEinside =

14

E0 43

P

P = 3

4 1 + 2

E0

E = 3 + 2

E0 and

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Bound Charge Consider a small area da inside a dielectric Polarization P = Nqs How many dipoles straddle this area?

Charge qNsda = Pda is split from the corresponding negative charge by the area da

Integrate this over a closed surface S How much (negative) charge remains inside?

Use the Divergence Theorem

P

da

s

Q = P da

S

da P

= divPBound charge distribution due to non-uniform polarization

Free and Bound Charges Inside dielectric, two types of charges may exist: Bound charge bound belongs to the dielectric material Appears only when E field polarizes the dielectric

Free charge free is brought in from outside

Both bound and free charges create E field

For a given free distribution and a constant

E follows Gausss Law with free except for a factor 1/

bound = divP P = 1

4E

divE = 4(free + bound)

divE =

4free

divE = 4free ( 1)divE

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Screening A point charge Q is inside a dielectric Q is the only free charge here

E field at distance r from the charge is

Polarization density P due to this field is

This creates bound charge density This div is zero everywhere except at the origin Integrate inside a very small sphere around Q

Polarization creates a screen around free charges

E

P

E = Q

r 2r

r

Coulomb reduced by 1/

P = 1

4E = ( 1)Q

4r 2r

bound = divP =

( 1)Q4

div rr 2

boundV dV = P daS =

1

Q Negative charge surrounds Q

Electric Displacement Electric displacement D is defined by For electric field inside an isotropic dielectric

D satisfies Gausss Law with free charge:

Importance of D is more historic than practical Its easy to calculate only in linear, isotropic dielectric In that case, writing D instead of E saves only a little ink

In a more complex medium, its safer to use E and 4P, and keep track of how the material reacts to E

D E + 4P

P = 1

4E D = E

divD = 4free

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Bound-Charge Current In a non-static system, P may vary with time Imagine s changing in response to changing E Charges +q and q move bound current

Jbound adds to the free current J in generating B

For a linear isotropic dielectric,

s

Jbound = N q

dr+dt

qdrdt

= Nq

dsdt

=Pt

B = 4

cJ+ P

t

+

1cEt

=4c

J+ 1c

t

E + 4P( )= D if you like

B = 4

cJ+

cEt

Electromagnetic Waves Inside a dielectric with no free charge and no free current

Same technique used in Lecture #18 turn them into

Solutions are waves propagating with speed

EM waves travel slower in a dielectric by factor n is the index of refraction of the material

E = 0 E = 1cBt

B = 0 B = cEt

2E =

c22Et 2

and 2B = c2

2Bt 2

c < c

n =

8

Electromagnetic Waves Plane wave solutions of the Maxwells eqns. are

Wave equation:

Divergence:

Curl:

Similar to the vacuum solutions except: Propagation velocity is reduced by |E| is smaller than |B| by the same factor

E = E0 sin(k r t) B = B0 sin(k r t)

2E =

c22Et 2

k 2 =c2

2

k=

c

E = 0 B = 0 k E0 k B0

E = 1

cBt

k E0 =c

B0 k E0 =

1

B0

1

Frequency Dependence Discussion so far applies to any dielectric = any insulator All insulators are transparent, with Index of refraction of water is

When E changes, it takes time for dielectrics to polarize Dielectric constant is constant only for static/slowly-changing field

Especially true for liquid of polar molecules, e.g. water Molecules must rotate Takes ~1011 seconds large up to 1010 Hz, then drops to an ordinary value of 1.78 Index of refraction of water for visible light is 1.33

80 = 8.9 n = Wrong!

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Ordinary Dielectrics Inside insulators are electrons bound to atoms They behave as mass-spring oscillators EM waves drive them with F = eE If the frequency is close to the resonance

frequency 0, the electrons oscillate strongly They absorb the incoming waves

Typical 0 for bound electrons are 10151016 Hz In the visible to ultraviolet (UV) region Loosely bound electrons (small k small 0) make material

opaque in visible light

Positively charged hydrogen atoms in molecules oscillate at lower 0 because of larger m Microwave is absorbed by water and organic compounds

e atom

light

E 0 =

kme

Summary Electric susceptibility of a dielectric At boundaries of dielectrics,

E|| is continuous, but E may not be If P is not uniform, bound charge appears

In such a medium, Maxwells equation is modified as:

Wave solutions propagate with a reduced speed Frequency dependence of makes the solutions more complex

P = eE =

14

E

bound = divP

divE = 4(free + bound) =

4free

for linear, isotropic dielectric

B = 4

cJ+

cEt

c