# Chapter 6 Electrostatic Boundary-Value Problems

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Chapter 6 Electrostatic Boundary-Value Problems
Lecture by Qiliang Li

Β§6.1 Introduction The goal is to determine electric field E
We need to know the charge distribution Use Coulombβs law π¬= ππ π πΉ 4π π 2 or π¬= ππ πΉ 4π π 3 Use Gaussβs law π= π π«βππΊ= π πππ Or We need to know the potential V π¬=βπ΅π

Β§6.1 Introduction However, we usually donβ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

Β§6.2 Poissonβs and Laplaceβs Equations
Poissonβs and Laplaceβs equations can be derived from Gaussβs law π»βπ·=π»βππΈ= π π Or π» 2 π=β π π π This is Poissonβs Eq. If π π =0, it becomes Laplaceβs Eq. π» 2 π=0

Continue 6.2 The Laplaceβs Eq. in different coordinates: π 2 π π π₯ 2 + π 2 π π π¦ 2 + π 2 π π π§ 2 =0 1 π π ππ π ππ ππ + 1 π 2 π 2 π π π 2 + π 2 π π π§ 2 =0 1 π 2 π ππ π 2 ππ ππ + 1 π 2 π πππ π ππ π πππ ππ ππ + 1 π 2 π ππ 2 π π 2 π π π 2 =0

Β§6.3 Uniqueness Theorem Uniqueness Theorem: If a solution to Laplaceβs equation can be found that satisfies the boundary conditions, then the solution is unique.

Β§6.4 General Procedures for Solving Poissonβs or Laplaceβs Equations
Solve Lβs Eq. or Pβ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 Apply BCs to determine V V ο  E ο  D ο  π±=ππ¬ Find Q induced on conductor π= π π ππ , where π π = π· π ο C=Q/V ο  πΌ= π½ππ ο R

Example 6.1 (page 219)

Example 6.2 Details in P222

Example 6.3

Β§6.5 Resistance and Capacitance
Rβs definition (for all cross sections) π= π πΌ = π¬βππ ππ¬βππΊ Procedure to calculate R: Choose a suitable coordinate system Assume V0 as the potential difference b/w two ends Solve π» 2 π=0 to obtain V, find E from π¬=βπ»π, then find I from I= ππ¬ βππΊ Finally, obtain π= π 0 /πΌ

Assuming Q and determining V in terms of Q (involving Gaussβs law)
(continue) Capacitance is the ratio of magnitude of the charge on one of the plates to the potential difference between them. πΆ= π π = π π¬βππΊ π¬βππ Two methods to find C: Assuming Q and determining V in terms of Q (involving Gaussβs law) Assuming V and determining Q interms of V (involving solving Laplaceβs Eq.)

First methods Qο V, procedure: Choose a suitable coordinate system
(continue) First methods Qο V, procedure: Choose a suitable coordinate system Let the two conductor plates carry Q and βQ Determine E by using Gaussβs law and find V from π=β π¬βππ . (Negative sign can be ignored. We are interested at absolute value of V) Finally, obtain C from Q/V

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(continue) Example 6.8: a metal bar of conductivity Ο 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 Ο=a and Ο=b is π= 2πππ/π πππ‘ (b) the resistance between the two horizontal surface at z=0 and z=t is π β² = 4π‘ ππ( π 2 β π 2 ) z x y b a

(a) Use Laplaceβs Eq. in cylindrical coordinate system: π» 2 π= 1 π π ππ π ππ ππ =0 ο π=π΄πππ+π΅ π π=π =0β0=π΄πππ+π΅ ππ π΅=βπ΄πππ π π=π = π 0 =π΄πππ+π΅ ππ π΄= π 0 ππ π π So, π=π΄πππβπ΄πππ= π 0 ππ π π ππ π π , π¬=βπ»π=β π 0 πππ π π π π , π½=ππ¬, ππΊ=βπππππ§ π π , πΌ= π±βππΊ= π 2 π‘ π 0 π ππ π π , π= π 0 πΌ = 2ππ π π πππ‘

(b) Use Laplaceβs Eq. in cylindrical coordinate system: π» 2 π= π 2 π π π§ 2 =0 ο π=π΄π§+π΅ π π§=0 =0β0=0+π΅ ππ π΅=0 π π=π‘ = π 0 =π΄π‘+π΅ ππ π΄= π 0 π‘ So, π= π 0 π‘ π§, π¬=βπ»π=β π 0 π‘ π π , π½=ππ¬, ππΊ=βπππππ π π , πΌ= π±βππΊ= π 0 ππ( π 2 β π 2 ) 4π‘ , R=V0/I=? (please also use conventional method for (b))

Example 6.9: a coaxial cable contains an insulating material of conductivity Ο. 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Ί=2ππ/ππ π π Solve: Let V(Ο=a)=0 and V(Ο=a)=V0 π½=ππ¬= βπ π 0 ππππ/π , ππΊ=βπππππ§ π π πΌ= 2ππΏπ π 0 πππ/π R per unit length =V/I/L G=1/R= β¦ a b

Example 6. 10: find the charge in shells and the capacitance
Example 6.10: find the charge in shells and the capacitance. Solve: Use Laplaceβs Eq. (spherical) And BCs, π= π 0 ( 1 π β 1 π ) ( 1 π β 1 π ) So E=-dV/dr ar, Q=?, C=Q/V0

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): πΆ= π π = ππ π Solve: From Laplaceβs Eq.: π» 2 π=0 π 2 π π₯ 2 π=0ο π=π΄π₯+π΅ From BCs: V(0)=0 and V(x=d)=V0 π= π 0 π π₯ So, πΈ=βπ»π=β π 0 π π π , the surface charge: π π  =π·β π π =β ππ 0 π ο π= π π  π= ππ 0 π S, so: πΆ= ππ π x d V0

Example 6.12: determine the capacitance of each of th Πr1=4 e capacitors in Figure Take Πr2=6, d=5mm, S=30 cm2. Solve: (do it by yourself) Πr1 d/2 Πr2 d/2 Πr1 Πr2 w/2 w/2

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 Πr=(10+Ο)/Ο, where Ο 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 βQ. π=β π π π 2π π 0 π π ππΏ ππ= π 2π π 0 ππ 10+π 10+π ο C=Q/V=β¦=434.6 pF/m

Β§6.6 Method of Images The method of images is introduced by Lord Kelvin to determine V, E and D, avoiding Poisonβ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.

Conducting plane grounded Image charge So that the potential at the plane position = 0 V

In applying the image method, two conditions must always be satisfied:
Equipotential V = 0 Perfect conducting surface grounded In applying the image method, two conditions must always be satisfied: The image charge(s) must be located in the conducting region (satify Poissonβs Eq.) The image charge(s) must be located such that on the conducting surface(s) the potential is zero or constant

A point charge above a grounded conducting plance