Electromagnetism Physics 15b Lecture #22 Maxwell’s Equations in Dielectrics
Purcell 10.11–10.15
What We Did Last Time 2p cos θ
A dipole generates electric field Er = r3 A dipole in an electric field receives: Torque N = p × E If E is non-uniform, net force F = (p ⋅ ∇)E
, Eθ =
p sin θ r3
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〉 = −4πP
A uniformly polarized sphere
External field looks like produced by a dipole VP Internal field E = −(4π/3)P
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Today’s Goals Connect polarization P and the dielectric constant ε 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 Maxwell’s equations Electromagnetic waves in dielectrics
Dielectric Constant How does P relate to the dielectric constant ε? Consider the filled capacitor again
−Q
+Q
Electric field is reduced by factor 1/ε
E=
4πσ = 4πσ − 4π P ε
Field from charge on the plates
Field from polarization
P ε −1 = ≡ χ e electric susceptibility E 4π d Polarization density P is related to the average electric field E that causes it by: ε −1 E = ε E − 4π P
This is an empirical law, and is “correct” within limits
P = χ eE =
4π
E
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Uniformly Polarized Sphere A dielectric sphere is uniformly polarized along +z
It contains dipoles p = qs with density N
P = Np = Nqs
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
R
Field outside is identical to that generated by a single dipole moment NqVs = V P We know this field V P ⋅ rˆ VP cos θ = from the last lecture: ϕ (r > R) = r2 r2
Uniformly Polarized Sphere Inside the sphere, there is no net charge The field must obey Laplace’s eqn.
We also need the boundary condition, i.e., values of φ at the surface, which we know from the outside field
ϕ (r = R) =
VP cos θ V 4π = 3 Pr cos θ = Pz 2 3 R R
A uniform electric field along +z works
ϕ (r < R) =
4π Pz 3
E(r < R) = −
4π P 3
The field due to a uniformly polarized sphere is Inside: E = −(4π/3)P Outside: identical to the field generated by a dipole (4π/3)R3P
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Boundary Conditions Electric potential φ 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: ⎧ ⎧ 8π 4π P cos θ P cos θ ⎪⎪Er = ⎪⎪Er = − 3 3 Outside ⎨ Inside ⎨ ⎪E = 4π P sin θ ⎪E = 4π P sin θ θ ⎪⎩ ⎪⎩ θ 3 3
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 4π P generates a uniform field inside: E′ = − P 3 4π Total field inside is Einside = E0 + E′ = E0 − P 3 Resulting polarization is ε − 1⎛ 4π ⎞ P = χ eEinside = E0 − P ⎜ 4π ⎝ 3 ⎟⎠
We can solve this to find ⎛ 3 ⎞ E′ = ⎜ E ⎝ ε + 2 ⎟⎠ 0
and
P=
3 ⎛ ε − 1⎞ E 4π ⎜⎝ ε + 2 ⎟⎠ 0
Uniform external field polarizes the sphere uniformly
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Bound Charge Consider a small area da inside a dielectric
da
Polarization P = Nqs How many dipoles “straddle” this area?
s
P
Charge qNs⋅da = P⋅da is split from the corresponding negative charge by the area da
Integrate this over a closed surface S
da
How much (negative) charge remains inside?
P
Q = − ∫ P ⋅ da S
Use the Divergence Theorem charge distribution = − div P “Bound” due to non-uniform polarization
ρ
Free and Bound Charges Inside dielectric, two “types” of charges may exist:
“Bound” charge ρbound belongs to the dielectric material Appears only when E field polarizes the dielectric
ρbound = − div P
P=
ε −1 E 4π
“Free” charge ρfree is brought in from outside
Both “bound” and “free” charges create E field div E = 4π ( ρfree + ρbound )
For a given ρfree distribution and a constant ε
4πρfree ε E follows Gauss’s Law with ρfree except for a factor 1/ε div E = 4πρfree − (ε − 1) div E
div E =
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Screening A point charge Q is inside a dielectric Q
E
is the only “free” charge here
E field at distance r from the charge is
r
P
Q rˆ Coulomb reduced by 1/ε εr 2 Polarization density P due to this field is ε −1 (ε − 1)Q P= E= rˆ 4π 4πε r 2 ⎛ rˆ ⎞ (ε − 1)Q div ⎜ 2 ⎟ This creates bound charge density ρbound = div P = 4πε ⎝r ⎠ This div is zero everywhere except at the origin Integrate inside a very small sphere around Q ε −1 ∫V ρbounddV = − ∫S P ⋅ da = − ε Q Negative charge surrounds Q E=
Polarization creates a “screen” around free charges
Electric Displacement Electric displacement D is defined by D ≡ E + 4π P
For electric field inside an isotropic dielectric ε −1 P= E D = εE 4π
D satisfies Gauss’s Law with free charge:
div D = 4πρfree
Importance of D is more historic than practical It’s easy to calculate only in linear, isotropic dielectric In that case, writing D instead of εE saves only a little ink In a more complex medium, it’s safer to use E and 4πP, and keep track of how the material reacts to E
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Bound-Charge Current In a non-static system, P may vary with time Imagine Charges
s changing in response to changing E +q and −q move “bound” current
s
⎛ dr dr ⎞ ds ∂P Jbound = N ⎜ q + − q − ⎟ = Nq = dt ⎠ dt ∂t ⎝ dt
Jbound adds to the “free” current J in generating B ∇×B = For
4π c
⎛ ∂P ⎞ 1 ∂E 4π 1 ∂ ⎜⎝ J + ∂t ⎟⎠ + c ∂t = c J + c ∂t E + 4π P
(
= D if you like
a linear isotropic dielectric,
∇×B =
)
4π ε ∂E J+ c c ∂t
Electromagnetic Waves Inside a dielectric with no free charge and no free current 1 ∂B ∇ ⋅E = 0 ∇ × E = − c ∂t ε ∂E ∇ ⋅B = 0 ∇ × B = c ∂t Same
