Presentation on theme: "Chapter 6 Electrostatic Boundary-Value Problems"— Presentation transcript:
1 Chapter 6 Electrostatic Boundary-Value Problems Lecture by Qiliang Li
2 §6.1 Introduction The goal is to determine electric field E We need to know the charge distributionUse Coulomb’s law𝑬= 𝑑𝑄 𝒂 𝑹 4𝜋 𝑅 2 or 𝑬= 𝑑𝑄 𝑹 4𝜋 𝑅 3Use Gauss’s law𝜓= 𝑆 𝑫∙𝑑𝑺= 𝑄 𝑒𝑛𝑐Or We need to know the potential V𝑬=−𝜵𝑉
3 §6.1 IntroductionHowever, we usually don’t know the charge distribution or potential profile inside the medium.In most cases, we can observe or measure the electrostatic charge or potential at some boundaries We can determine the electric field E by using the electrostatic boundary conditions
4 §6.2 Poisson’s and Laplace’s Equations Poisson’s and Laplace’s equations can be derived from Gauss’s law𝛻∙𝐷=𝛻∙𝜖𝐸= 𝜌 𝑉Or𝛻 2 𝑉=− 𝜌 𝑉 𝜖This is Poisson’s Eq.If 𝜌 𝑉 =0, it becomes Laplace’s Eq.𝛻 2 𝑉=0
6 §6.3 Uniqueness TheoremUniqueness Theorem: If a solution to Laplace’s equation can be found that satisfies the boundary conditions, then the solution is unique.
7 §6.4 General Procedures for Solving Poisson’s or Laplace’s Equations Solve L’s Eq. or P’s Eq. using (a) direct integration when V is a function of one variable or (b) separation of variables if otherwise. solution with constants to be determinedApply BCs to determine VV E D 𝑱=𝜎𝑬Find Q induced on conductor 𝑄= 𝜌 𝑆 𝑑𝑆 , where 𝜌 𝑆 = 𝐷 𝑛 C=Q/V 𝐼= 𝐽𝑑𝑆 R
16 §6.5 Resistance and Capacitance R’s definition (for all cross sections)𝑅= 𝑉 𝐼 = 𝑬∙𝑑𝒍 𝜎𝑬∙𝑑𝑺Procedure to calculate R:Choose a suitable coordinate systemAssume V0 as the potential difference b/w two endsSolve 𝛻 2 𝑉=0 to obtain V, find E from 𝑬=−𝛻𝑉, then find I from I= 𝜎𝑬 ∙𝑑𝑺Finally, obtain 𝑅= 𝑉 0 /𝐼
17 Assuming Q and determining V in terms of Q (involving Gauss’s law) (continue)Capacitance is the ratio of magnitude of the charge on one of the plates to the potential difference between them.𝐶= 𝑄 𝑉 = 𝜖 𝑬∙𝑑𝑺 𝑬∙𝑑𝒍Two methods to find C:Assuming Q and determining V in terms of Q (involving Gauss’s law)Assuming V and determining Q interms of V (involving solving Laplace’s Eq.)
18 First methods QV, procedure: Choose a suitable coordinate system (continue)First methods QV, procedure:Choose a suitable coordinate systemLet the two conductor plates carry Q and –QDetermine E by using Gauss’s law and find V from 𝑉=− 𝑬∙𝑑𝒍 . (Negative sign can be ignored. We are interested at absolute value of V)Finally, obtain C from Q/V
23 (continue)Example 6.8: a metal bar of conductivity σ is bent to form a flat 90o sector of inner radius a, outer radius b, and thickness t as shown in Figure Show that (a) the resistance of the bar between the vertical curved surfaces at ρ=a and ρ=b is 𝑅= 2𝑙𝑛𝑏/𝑎 𝜎𝜋𝑡 (b) the resistance between the two horizontal surface at z=0 and z=t is 𝑅 ′ = 4𝑡 𝜎𝜋( 𝑏 2 − 𝑎 2 )zxyba
26 Example 6.9: a coaxial cable contains an insulating material of conductivity σ. If the radius of the central wire is a and that of the sheath is b, show that the conductance of the cable per unit length is 𝐺=2𝜋𝜎/𝑙𝑛 𝑏 𝑎 Solve: Let V(ρ=a)=0 and V(ρ=a)=V0 𝐽=𝜎𝑬= −𝜎 𝑉 0 𝜌𝑙𝑛𝑏/𝑎 , 𝑑𝑺=−𝜌𝑑𝜙𝑑𝑧 𝒂 𝝆 𝐼= 2𝜋𝐿𝜎 𝑉 0 𝑙𝑛𝑏/𝑎 R per unit length =V/I/L G=1/R= …ab
27 Example 6. 10: find the charge in shells and the capacitance Example 6.10: find the charge in shells and the capacitance. Solve: Use Laplace’s Eq. (spherical) And BCs, 𝑉= 𝑉 0 ( 1 𝑟 − 1 𝑏 ) ( 1 𝑎 − 1 𝑏 ) So E=-dV/dr ar, Q=?, C=Q/V0
28 Example 6. 11: assuming V and finding Q to derive Eq. (6 Example 6.11: assuming V and finding Q to derive Eq. (6.22): 𝐶= 𝑄 𝑉 = 𝜖𝑆 𝑑 Solve: From Laplace’s Eq.: 𝛻 2 𝑉=0 𝑑 2 𝑑 𝑥 2 𝑉=0𝑉=𝐴𝑥+𝐵 From BCs: V(0)=0 and V(x=d)=V0 𝑉= 𝑉 0 𝑑 𝑥 So, 𝐸=−𝛻𝑉=− 𝑉 0 𝑑 𝒂 𝒙 , the surface charge: 𝜌 𝑠 =𝐷∙ 𝒂 𝒏 =− 𝜖𝑉 0 𝑑 𝑄= 𝜌 𝑠 𝑆= 𝜖𝑉 0 𝑑 S, so: 𝐶= 𝜖𝑆 𝑑xdV0
29 Example 6.12: determine the capacitance of each of th Єr1=4 e capacitors in Figure Take Єr2=6, d=5mm, S=30 cm2. Solve: (do it by yourself)Єr1d/2Єr2d/2Єr1Єr2w/2w/2
30 Example 6. 13: A cylindrical capacitor has radii a=1cm and b=2. 5cm Example 6.13: A cylindrical capacitor has radii a=1cm and b=2.5cm. If the space between the plates is filled with and inhomogeneous dielectric with Єr=(10+ρ)/ρ, where ρ is in centimeters, find the capacitance per meter of the capacitor. Solve: Use Eq. 6.27a, set the inner shell with +Q and outer shell with –Q. 𝑉=− 𝑏 𝑎 𝑄 2𝜋 𝜖 0 𝜖 𝑟 𝜌𝐿 𝑑𝜌= 𝑄 2𝜋 𝜖 0 𝑙𝑛 10+𝑏 10+𝑎 C=Q/V=…=434.6 pF/m
31 §6.6 Method of ImagesThe method of images is introduced by Lord Kelvin to determine V, E and D, avoiding Poison’s Eq.Image Theory states that a given charge configuration above an infinite grounded perfect conducting plane may be replaced by the charge configuration itself, its image, and an equipotential surface in place of the conducting plane.
32 Conducting planegroundedImage chargeSo that the potential at the plane position = 0 V
33 In applying the image method, two conditions must always be satisfied: Equipotential V = 0Perfect conducting surface groundedIn applying the image method, two conditions must always be satisfied:The image charge(s) must be located in the conducting region (satify Poisson’s Eq.)The image charge(s) must be located such that on the conducting surface(s) the potential is zero or constant
34 A point charge above a grounded conducting plance
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