Chapter 6 Electrostatic Boundary-Value Problems Lecture by Qiliang Li 1 §6.1 Introduction The goal is to determine electric field E • We need to know the charge distribution – Use Coulomb’s law π¬= ππ ππΉ 4ππ 2 or π¬ = – Use Gauss’s law π= π ππ πΉ 4ππ 3 π« β ππΊ = ππππ • Or We need to know the potential V π¬ = −π΅π 2 §6.1 Introduction • However, 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 ο¨ We can determine the electric field E by using the electrostatic boundary conditions 3 §6.2 Poisson’s and Laplace’s Equations • Poisson’s and Laplace’s equations can be derived from Gauss’s law π» β π· = π» β ππΈ = ππ Or ππ 2 π» π=− π This is Poisson’s Eq. If ππ = 0, it becomes Laplace’s Eq. π»2π = 0 4 Continue 6.2 The Laplace’s Eq. in different coordinates: π2π π2π π2π + 2+ 2 =0 2 ππ₯ ππ¦ ππ§ 1 π ππ 1 π2π π2π π + 2 + 2 =0 2 π ππ ππ π ππ ππ§ 1 π ππ 1 π ππ 2 π + 2 π πππ 2 π ππ ππ π π πππ ππ ππ 1 π2π + 2 2 =0 2 π π ππ π ππ 5 §6.3 Uniqueness Theorem Uniqueness Theorem: If a solution to Laplace’s equation can be found that satisfies the boundary conditions, then the solution is unique. 6 §6.4 General Procedures for Solving Poisson’s or Laplace’s Equations 1. 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. ο solution with constants to be determined 2. Apply BCs to determine V 3. V ο E ο D ο π± = ππ¬ 4. Find Q induced on conductor π = ππ ππ, where ππ = π·π ο C=Q/V ο πΌ = π½ππο R 7 Example 6.1 (page 219) 8 9 10 Example 6.2 Details in P222 Example 6.3 13 14 15 §6.5 Resistance and Capacitance R’s definition (for all cross sections) π¬ β ππ π π = = πΌ ππ¬ β ππΊ Procedure to calculate R: 1. Choose a suitable coordinate system 2. Assume V0 as the potential difference b/w two ends 3. Solve π» 2 π = 0 to obtain V, find E from π¬ = −π»π, then find I from I = ππ¬ β ππΊ 4. Finally, obtain π = π0 /πΌ 16 (continue) Capacitance is the ratio of magnitude of the charge on one of the plates to the potential difference between them. π π π¬ β ππΊ πΆ= = π π¬ β ππ Two methods to find C: 1. Assuming Q and determining V in terms of Q (involving Gauss’s law) 2. Assuming V and determining Q interms of V (involving solving Laplace’s Eq.) 17 (continue) First methods Qο V, procedure: 1. Choose a suitable coordinate system 2. Let the two conductor plates carry Q and –Q 3. Determine E by using Gauss’s law and find V from π = − π¬ β ππ. (Negative sign can be ignored. We are interested at absolute value of V) 4. Finally, obtain C from Q/V 18 (continue) 19 (continue) 20 (continue) 21 (continue) 22 (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 6.17. Show that (a) the resistance of the bar between the vertical curved surfaces at ρ=a 2πππ/π and ρ=b is π = πππ‘ (b) the resistance between the two horizontal surface at z=0 and z=t is 4π‘ ′ π = ππ(π 2 − π2 ) x z b a y 23 (a) Use Laplace’s Eq. in cylindrical coordinate system: 1 π ππ 2 π» π= π =0 π ππ ππ ο π = π΄πππ + π΅ π π = π = 0 → 0 = π΄πππ + π΅ ππ π΅ = −π΄πππ π0 π π = π = π0 = π΄πππ + π΅ ππ π΄ = π ππ π π0 π So, π = π΄πππ − π΄πππ = π ππ , ππ π¬ = −π»π = − πΌ= π± β ππΊ = π0 π πππ π π π ππ , π½ = ππ¬, ππΊ = −πππππ§ππ , π π‘π0 π , 2 ππ π π π = π0 πΌ = 2ππ π π πππ‘ 24 (b) Use Laplace’s Eq. in cylindrical coordinate system: 2 π π 2 π» π= 2 =0 ππ§ ο π = π΄π§ + π΅ π π§ = 0 = 0 → 0 = 0 + π΅ ππ π΅ = 0 π0 π π = π‘ = π0 = π΄π‘ + π΅ ππ π΄ = π‘ π0 So, π = π§, π‘ π¬= πΌ= π0 −π»π = − ππ , π½ = ππ¬, ππΊ = −πππππππ , π‘ π0 ππ(π2 −π2 ) π± β ππΊ = , R=V0/I=? 4π‘ (please also use conventional method for (b)) 25 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 π πΊ = 2ππ/ππ π Solve: Let V(ρ=a)=0 and V(ρ=a)=V0 π½ = ππ¬ = πΌ= 2ππΏππ0 πππ/π −ππ0 , ππππ/π ππΊ = −πππππ§ππ R per unit length =V/I/L G=1/R= … a b 26 Example 6.10: find the charge in shells and the capacitance. Solve: Use Laplace’s Eq. (spherical) And BCs, π = π0 1 1 ( − ) π π 1 1 ( − ) π π V0=100V b a GND Πr So E=-dV/dr ar, Q=?, C=Q/V0 27 Example 6.11: assuming V and finding Q to derive π ππ Eq. (6.22): πΆ = = π π Solve: From Laplace’s Eq.:π» 2 π = 0 π2 π 2 ππ₯ = 0ο π = π΄π₯ + π΅ x d V0 0 0 From BCs: V(0)=0 and V(x=d)=V0 π0 π= π₯ π π0 So, πΈ = −π»π = − ππ , the surface charge: ππ = π· β ππ = ππ0 − ο π π π = ππ π ππ0 = S, π so: πΆ = ππ π 28 Example 6.12: determine the capacitance of each of th Πr1=4 e capacitors in Figure 6.20. Take Πr2=6, d=5mm, S=30 cm2. Solve: d/2 Π (do it by yourself) Π d/2 r1 r2 Πr1 Πr2 w/2 w/2 29 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 Πr=(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 π π outer shell with –Q. π = − π ππ = π 10+π ππ 2ππ0 10+π 2ππ0 ππ ππΏ ο C=Q/V=…=434.6 pF/m 30 §6.6 Method of Images • The 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. 31 Conducting plane grounded Image charge So that the potential at the plane position = 0V 32 Equipotential V=0 Perfect conducting surface grounded In applying the image method, two conditions must always be satisfied: 1. The image charge(s) must be located in the conducting region (satify Poisson’s Eq.) 2. The image charge(s) must be located such that on the conducting surface(s) the potential is zero or constant 33 A. A point charge above a grounded conducting plance 34