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Transcript
Electromagnetism Physics 15b Lecture #20 Dielectrics Electric Dipoles Purcell 10.1–10.6 What We Did Last Time Plane wave solutions of Maxwell’s equations ⎧⎪E = E0 sin(k ⋅ r − ω t) ω = kc, E0 = B0 , k̂ = Ê × B̂ ⎨ ⎩⎪B = B0 sin(k ⋅ r − ω t) Propagates along k with speed c Two possible polarizations for the same k Energy flow given by the Poynting vector S = For plane waves, average energy flow is S = c E×B 4π c 2 c 2 E k̂ = E k̂ 4π 8π 0 1 Goals Today Introduce dielectric material Synonym for “insulator” Effects on electric field expressed by dielectric constant Universally found in capacitors Look into the microscopic origin of dielectric Dipole moment of small charge distribution Electric field generated by a dipole moment Force on a dipole moment due to electric field Next lecture: construct dielectric from dipoles Material in a Uniform E Field A slab of insulator is in a uniform E field E + and − charges feel the force F = qE F They can’t flow, but they do move slightly from the natural (equilibrium) positions Excess + and − charges appear on the surfaces Additional electric field E′ appears inside E ′ = 4πσ and opposite to E It’s reasonable to assume σ ∝ E Total electric field inside the insulator is 1 Ein = Eout + E′ = (1− k)Eout = Eout ε ε = the dielectric constant of the material +σ E′ −σ ε > 1, i.e., the electric field is weaker inside 2 Dielectric Constant Dielectric constant ε depends on the density Gases very close to 1 Liquids and solids 2 to 10 Molecules can rotate to align with electric field Prime example: water + − O − H − O − H + Condition ε Air Gas, 0°C 1.0006 Methane Gas, 0°C 1.0009 HCl Gas, 0°C 1.0046 Gas, 110°C 1.0126 Water Liquids and gases of polar molecules have large ε + H Substance E H + See textbook §10.6 and 10.12 Liquid, 20°C 80.4 Benzene Liquid, 20°C 2.28 Methanol Liquid, 20°C 33.6 Ammonia Liquid, −34°C 22.6 Mineral oil Liquid, 20°C 2.24 NaCl Solid, 20°C 6.12 Sulfur Solid, 20°C 4.0 Silicon Solid, 20°C 11.7 Polyethylene Solid, 20°C 2.25–2.3 Porcelain 6.0–8.0 Solid, 20°C Dielectric in Capacitor Capacitor is filled with a dielectric −Q +Q 4π Q area A Charge ±Q on the plates create E0 = A Dielectric polarizes and creates surface charges E 4π Q Actual electric field inside is E = 0 = ε εA Potential difference is V= ∫ right left E ⋅ds = 4π Qd εA C= Q εA = = ε ⋅C0 V 4π d Dielectric increases the capacitance by a factor ε d capacitance w/o dielectric Real-world capacitors use various dielectric materials paper, ceramics, mica, oil, liquid electrolyte, etc. Issues: field (voltage) tolerance, frequency response, temperature dependence, polarity, long-term stability 3 Half-Filled Capacitor A rectangular capacitor is partially filled with a dielectric If it were empty, the capacitance would be ab C0 = 4π d Consider it as two capacitors: a(b − x) 4π d ε ax = 4π d Empty part: Cempty = Filled part: Cfilled a ε x d b Ctotal = a{b + (ε − 1)x} 4π d connected in parallel Suppose there is charge Q in this capacitor dU 2π dQ 2 (ε − 1) Q2 2π dQ 2 =− <0 = dx 2C a{b + (ε − 1)x} a{b + (ε − 1)x} 2 Increasing x decreases the potential energy Energy is U = Electrostatic force pulls the dielectric into the capacitor Small Charge Distribution So far we had uniform E field causing uniform polarization More general case (non-uniform E) requires a better framework Consider an arbitrary charge distribution of a small size, and the electric field far away from it A Electric potential at point A is ρ(r ′)dv ′ Integral over volume where ρ ≠ 0 ϕ=∫ V r − r′ Since r′ << r, we use Taylor expansion r − r′ −1 = (r − 2rr ′ cos θ + r ′ ) 2 2 −1 2 E r ρ≠0 r′ ⎞ 1⎛ r′ r ′ 2 (3 cos2 θ − 1) = ⎜ 1+ cos θ + 2 + ⎟ r⎝ r 2 r ⎠ 4 Moments We can now express the potential at A as ϕ= 1 1 ρ(r ′) dv ′ + 2 ∫ r V r K0 ∫ V ρ(r ′)r ′ cos θ dv ′ + 1 r3 ∫ V ρ(r ′)r ′ 2 (3 cos2 θ − 1) dv ′ + 2 K1 K2 /r2 At large distance, K0/r >> K1 >> K2 >> … We must consider higher-order terms only if the preceding terms happen to be zero /r3 K0 is the net charge of the source K0/r is the familiar Coulomb potential For a small charged object, the Coulomb force due to the net charge outweighs everything else If the object (e.g. a molecule) is net neutral, the K1 term becomes important Dipole Moment We can rewrite the K1 term as K1 r 2 = 1 r2 ∫ V r̂ ⋅ ρ(r ′)r ′ dv ′ r 2 ∫V ρ(r ′)r ′ cos θ dv ′ = a vector determined by the charge distribution Define the dipole moment of a charge distribution by p≡ ∫ V ρ(r ′)r ′ dv ′ then the electric potential due to it is ϕ= r̂ ⋅ p r2 Q: Doesn’t this definition depend on where the origin of the coordinate system is? A: It does. But that’s OK 5 Coordinate Origin Let’s move the origin by Δr It’s still inside the charge distribution r → r − Δr r ′ → r ′ − Δr The dipole moment becomes p → ∫ ρ(r ′)(r ′ − Δr)dv ′ = p − QΔr Δr V A E r r′ If Q = 0, no change to p If Q ≠ 0, we now have a different p Re-calculate the first two terms of the potential ϕ= Q r̂ ⋅ p Q (r − Δr) ⋅ (p − QΔr) + 2 → + 3 r r r − Δr r − Δr Taylor expand by Δr/r and take the leading terms ⎛ 1 r̂ ⋅ Δr ⎞ r̂ ⋅ (p − QΔr) Q ⎜ + 2 + ⎟ + + ⎝r ⎠ r r2 Extra terms cancel between the Coulomb and the dipole parts Simple Electric Dipole Two charges, +q and –q, separated by a distance s p≡ ∫ V This is a useful model for any neutral object (e.g. a molecule) with a dipole moment p s ρ(r ′)r ′ dv ′ = +qs − q0 = qs −q +q Electric field at large distances will be identical We know how to draw field lines around this 6 Dipole Electric Field ⎛ r̂ ⋅ p ⎞ The electric field due to a dipole p is E = −∇ϕ = −∇ ⎜ 2 ⎟ ⎝ r ⎠ Use spherical coordinates, the polar axis parallel to p ϕ= p cos θ r2 ⎧ ∂ϕ 2p cos θ = ⎪⎪Er = − ∂r r3 ⎨ ⎪E = − 1 ∂ϕ = p sin θ ⎪⎩ θ r ∂θ r3 p E field decreases with 1/r3 Faster than Coulomb, as expected Force on Dipole An electric dipole p is in an external field E Force qE acts on each charge q Net force is zero if total charge Q is zero +q +qE Forces on +q and –q may be equal and opposite, but not on the same line s E −qE −q Combined, they produce a torque N = ∑ r × F = r+ × qE + r− × (−qE) = q(r+ − r− ) × E = qs × E = p × E The torque N rotates the dipole so that it will line up with E Also: the net force F might not be zero if E is not uniform F = qE(r+ ) − qE(r− ) = q(s ⋅ ∇)E = (p ⋅ ∇)E This one is tricky, so let’s look at a simple example 7 Dipole and a Charge A dipole p is near a charge Q Assume s << r p is lined up with E Q −qE −q s +q +qE r Net force is ⎛ Q Q⎞ 2qQ F = qE(r + s) − qE(r) = q ⎜ − ≈− 3 2 2⎟ ⎝ (r + s) r ⎠ r i.e., the dipole is attracted to the charge What if Q is negative? Torque will rotate p so that it points toward the charge Net force F will point toward the charge again Electric dipole is generally pulled toward stronger E field This is why neutral objects (e.g. dust particles) are attracted by static electricity Summary 1 ε Electric field inside dielectric is reduced Ein = Eout ε = dielectric constant Capacitance is increased by factor ε Far field due to small charge distribution is determined by: First, the net charge Q Coulomb field If Q = 0, then the dipole moment p ≡ ρ(r ′)r ′ dv ′ V s Modeled by ∫ p = qs −q ϕ= r̂ ⋅ p r2 +q A dipole in an electric field receives: Torque N = p × E If E is non-uniform, net force F = (p ⋅ ∇)E Dipoles are attracted to stronger E field 8