Electromagnetics: From WirelessELEC 3600 to Photonic Applications Tutorial 5 Electrostatic Boundary-value Problems Aaron, Chenxiang ZHAO [email protected], Rm 3121A Department of Electronic & Computer Engineering HKUST Spring 2014 Outline Poisson’s and Laplace’s equations Capacitance Parallel-plate capacitor Coaxial capacitor Spherical capacitor ELEC 3600 Electromagnetics: From Wireless to Photonic Applications Tutorial 5 Spring 2014 2 . Poisson’s and Laplace’s equations Poisson’s equation v V 2 Laplace’s equation v 0 V 0 2 Uniqueness theorem: Laplacian operator ELEC 3600 Electromagnetics: From Wireless to Photonic Applications Tutorial 5 Spring 2014 2 If a solution to Laplace’s equation can be found that satisfies the boundary conditions. then the solution is unique 3 . (a)True (b)False Q.Example 1 Q. equation 𝜕 2 V 1 𝜕V 𝜕 2 V + + + 10 = 0 𝜕𝜌2 𝜌 𝜕𝜌 𝜕𝑧 2 is called (a)Maxwell’s equation (d)Helmholtz’s equation (b)Laplace’s equation (e)Lorentz’s equation (c)Poisson’s equation ELEC 3600 Electromagnetics: From Wireless to Photonic Applications Tutorial 5 Spring 2014 4 .2 In cylindrical coordinates.1 Equation 𝛻 ∙ −𝜀𝛻𝑉 = 𝜌𝑉 may be regarded as Poisson's equation for an inhomogeneous medium. Capacitance Capacitor: two (or more) conductors carrying equal but opposite charges Definition of capacitance: Q E dS C V E dl To obtain C for any given two-conductor capacitance Method 1: Assuming Q and determining V in terms of Q (involving Gauss’s law) Method 2: Assuming V and determining Q in terms of V (involving solving Laplace’s equation) ELEC 3600 Electromagnetics: From Wireless to Photonic Applications Tutorial 5 Spring 2014 5 . Capacitance Parallel-plate capacitor Coaxial capacitor Spherical capacitor ELEC 3600 Electromagnetics: From Wireless to Photonic Applications Tutorial 5 Spring 2014 Q S C V d Q 2 L C b V ln a Q 4 C V 11 a b 6 . Example 2 Consider an isolated parallel-plate capacitor that each of the plates has an area S and they are separated by a distance d. The amount of uniformly distributed charges on each plate is Q. How will the following parameters change if the separated distance d of parallel plates increases (Q remains the same)? +Q (a) Potential difference V (b) Electric field E -Q (c) Energy stored in the capacitor WE Solution: (a) increase (b) no change (c) increase ELEC 3600 Electromagnetics: From Wireless to Photonic Applications Tutorial 5 Spring 2014 7 . (Assume the distance between the plates is much smaller than the size of the plates) +Q Solution: Q2 F 2 S ELEC 3600 Electromagnetics: From Wireless to Photonic Applications Tutorial 5 Spring 2014 -Q 8 . +Q and –Q are uniformly distributed on the plates. The separation distance is d and the area of each plate is S.Example 3 Determine the force between the plates of a parallelplate capacitor. If V(𝜌 = 5 mm) = 100 V and V(𝜌 = 15 mm) = 0 V.Example 4 The cylindrical capacitor whose cross section is shown in figure below has inner and outer radii of 5 mm and 15 mm.0) ELEC 3600 Electromagnetics: From Wireless to Photonic Applications Tutorial 5 Spring 2014 9 . calculate V. E. (Take 𝜀r = 2. respectively. and D at 𝜌 = 10 mm and 𝜌s on each plate. 𝑉 𝜌 = 𝑎 = 0 → 0 = 𝐴𝑙𝑛𝑎 + 𝐵 𝑜𝑟 𝐵 = −𝐴𝑙𝑛𝑎 𝑎 𝑉 𝜌 = 𝑏 = 𝑉0 → 𝑉0 = 𝐴𝑙𝑛𝑏 + 𝐵 = 𝐴𝑙𝑛𝑏 − 𝐴𝑙𝑛𝑎 = −𝐴𝑙𝑛 𝑏 𝑉0 𝑜𝑟 𝐴 = − 𝑎 𝑙𝑛 𝑏 ELEC 3600 Electromagnetics: From Wireless to Photonic Applications Tutorial 5 Spring 2014 10 . this becomes 𝑑𝑉 𝜌 =𝐴 𝑑𝜌 Or 𝑑𝑉 𝐴 = 𝑑𝜌 𝜌 Integrating one again yields 𝑉 = 𝐴𝑙𝑛𝜌 + 𝐵 Where A and B are constants of integration to be determined from the boundary conditions. upon multiplying by 𝜌 and integrating once.Example 4-Solution We solve for V in Laplace’s equation 𝛻 2 𝑉 = 0 in cylindrical coordinates. 1 𝑑 𝑑𝑉 𝛻2𝑉 = 𝜌 =0 𝜌 𝑑𝜌 𝑑𝜌 As 𝜌=0 is excluded. 𝑉 = 10 5 15 𝑙𝑛 5 100𝑙𝑛 = 36.Example 4-Solution 𝜌 𝑎 𝑉 = 𝐴𝑙𝑛𝜌 − 𝐴𝑙𝑛𝑎 = 𝐴𝑙𝑛 = − Hence. V0=100V.91𝑉.3𝑛𝐶/𝑚2 36𝜋 15𝑙𝑛3 𝑫 = 9. a=15mm.102𝒂𝝆 𝑉/𝑚 10 2𝒂𝝆 = 161𝒂𝝆 𝑛𝐶/𝑚2 36𝜋 −9 10 105 𝜌𝑆 𝜌 = 15𝑚𝑚 = − 2 = −107. 𝜀r=2. 𝜌𝑆 𝜌 = 𝑏 = 𝑫 ∙ 𝒂𝝆 = 𝐷𝑛 = At 𝜌=10mm.102 × 103 × 10−9 105 𝜌𝑆 𝜌 = 5𝑚𝑚 = 2 = 322𝑛𝐶/𝑚2 36𝜋 5𝑙𝑛3 ELEC 3600 Electromagnetics: From Wireless to Photonic Applications Tutorial 5 Spring 2014 11 . b=5mm. 𝑬 = −𝛻𝑉 = 𝑉0 𝑎 𝒂𝝆 𝜌𝑙𝑛 𝑏 𝑉0 𝑎 𝑙𝑛 𝑏 𝑙𝑛 𝑫 = 𝜀𝐸 = 𝜌 𝑎 𝜀0 𝜀𝑟 𝑉0 𝑎 𝒂𝝆 𝜌𝑙𝑛 𝑏 𝜀0 𝜀𝑟 𝑉0 𝑎 𝜌=𝑏 𝜌𝑙𝑛 𝑏 𝜀0 𝜀𝑟 𝑉0 𝜌𝑆 𝜌 = 𝑎 = 𝑫 ∙ −𝒂𝝆 = −𝐷𝑛 = − 𝑎 𝜌=𝑎 𝜌𝑙𝑛 𝑏 In this case. 𝑬 = 100 𝒂 10×10−3 𝑙𝑛3 𝝆 −9 = 9.