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**Chapter 6 Electrostatic Boundary-Value Problems**

Lecture by Qiliang Li

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**Β§6.1 Introduction The goal is to determine electric field E**

We need to know the charge distribution Use Coulombβs law π¬= ππ π πΉ 4π π
2 or π¬= ππ πΉ 4π π
3 Use Gaussβs law π= π π«βππΊ= π πππ Or We need to know the potential V π¬=βπ΅π

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Β§6.1 Introduction However, we usually donβt 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

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**Β§6.2 Poissonβs and Laplaceβs Equations**

Poissonβs and Laplaceβs equations can be derived from Gaussβs law π»βπ·=π»βππΈ= π π Or π» 2 π=β π π π This is Poissonβs Eq. If π π =0, it becomes Laplaceβs Eq. π» 2 π=0

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Continue 6.2 The Laplaceβs 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 ππ π ππ ππ + 1 π 2 π 2 π π π 2 + π 2 π π π§ 2 =0 1 π 2 π ππ π 2 ππ ππ + 1 π 2 π πππ π ππ π πππ ππ ππ + 1 π 2 π ππ 2 π π 2 π π π 2 =0

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

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**Β§6.4 General Procedures for Solving Poissonβs or Laplaceβs Equations**

Solve Lβs Eq. or Pβs 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±=ππ¬ Find Q induced on conductor π= π π ππ , where π π = π· π ο C=Q/V ο πΌ= π½ππ ο R

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Example 6.1 (page 219)

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Example 6.2 Details in P222

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Example 6.3

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**Β§6.5 Resistance and Capacitance**

Rβs definition (for all cross sections) π
= π πΌ = π¬βππ ππ¬βππΊ 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¬=βπ»π, then find I from I= ππ¬ βππΊ Finally, obtain π
= π 0 /πΌ

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**Assuming Q and determining V in terms of Q (involving Gaussβs law) **

(continue) Capacitance is the ratio of magnitude of the charge on one of the plates to the potential difference between them. πΆ= π π = π π¬βππΊ π¬βππ Two methods to find C: Assuming Q and determining V in terms of Q (involving Gaussβs law) Assuming V and determining Q interms of V (involving solving Laplaceβs Eq.)

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**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 βQ Determine E by using Gaussβs law and find V from π=β π¬βππ . (Negative sign can be ignored. We are interested at absolute value of V) Finally, obtain C from Q/V

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(continue)

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(continue)

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(continue)

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(continue)

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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
= 2πππ/π πππ‘ (b) the resistance between the two horizontal surface at z=0 and z=t is π
β² = 4π‘ ππ( π 2 β π 2 ) z x y b a

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(a) Use Laplaceβs Eq. in cylindrical coordinate system: π» 2 π= 1 π π ππ π ππ ππ =0 ο π=π΄πππ+π΅ π π=π =0β0=π΄πππ+π΅ ππ π΅=βπ΄πππ π π=π = π 0 =π΄πππ+π΅ ππ π΄= π 0 ππ π π So, π=π΄πππβπ΄πππ= π 0 ππ π π ππ π π , π¬=βπ»π=β π 0 πππ π π π π , π½=ππ¬, ππΊ=βπππππ§ π π , πΌ= π±βππΊ= π 2 π‘ π 0 π ππ π π , π
= π 0 πΌ = 2ππ π π πππ‘

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(b) Use Laplaceβs Eq. in cylindrical coordinate system: π» 2 π= π 2 π π π§ 2 =0 ο π=π΄π§+π΅ π π§=0 =0β0=0+π΅ ππ π΅=0 π π=π‘ = π 0 =π΄π‘+π΅ ππ π΄= π 0 π‘ So, π= π 0 π‘ π§, π¬=βπ»π=β π 0 π‘ π π , π½=ππ¬, ππΊ=βπππππ π π , πΌ= π±βππΊ= π 0 ππ( π 2 β π 2 ) 4π‘ , R=V0/I=? (please also use conventional method for (b))

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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 πΊ=2ππ/ππ π π Solve: Let V(Ο=a)=0 and V(Ο=a)=V0 π½=ππ¬= βπ π 0 ππππ/π , ππΊ=βπππππ§ π π πΌ= 2ππΏπ π 0 πππ/π R per unit length =V/I/L G=1/R= β¦ a b

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**Example 6. 10: find the charge in shells and the capacitance**

Example 6.10: find the charge in shells and the capacitance. Solve: Use Laplaceβs 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

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**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): πΆ= π π = ππ π Solve: From Laplaceβs 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, πΈ=βπ»π=β π 0 π π π , the surface charge: π π =π·β π π =β ππ 0 π ο π= π π π= ππ 0 π S, so: πΆ= ππ π x d V0

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Example 6.12: determine the capacitance of each of th Πr1=4 e capacitors in Figure Take Πr2=6, d=5mm, S=30 cm2. Solve: (do it by yourself) Πr1 d/2 Πr2 d/2 Πr1 Πr2 w/2 w/2

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**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 Πr=(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 βQ. π=β π π π 2π π 0 π π ππΏ ππ= π 2π π 0 ππ 10+π 10+π ο C=Q/V=β¦=434.6 pF/m

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

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

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**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 Poissonβs Eq.) The image charge(s) must be located such that on the conducting surface(s) the potential is zero or constant

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**A point charge above a grounded conducting plance**

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