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EE204B_Lesson_11_Poisson_and_Laplace_ - &euml;&sup3;&micro;&igrave;‚&not;&euml;&sup3;&cedil;

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Unformatted text preview: EE204B EE204B Electromagnetics Electromagnetics Lesson 11 Electrostatic BoundaryValue Problems Poisson’s and Laplace’s Equation Instructor: Prof. Yong H. Won Lesson 11 Topics Electrostatic Boundary-Value Problems (Chap. 6) • Poisson’s and Laplace’s Equations Eq • Uniqueness Theorem • Solution of Poisson’s and Laplace’s Equations EE204B Electromagnetics Prof. Yong H. Won Spring 2010 2 Lesson 11 Poisson’s and Laplace’s Equations So far we learned Electrostatics in free space and in material space. Given a certain charge distribution, we know how to find E by using either Coulomb’s law or Gauss’s law. However, in reality, often we don’t know charge distribution. Consider the following problem. z 1V z=d z=0 The upper plate at z = d is attached to 1 V. The lower plate at z = 0 is attached to ground What is E between the two plates? EE204B Electromagnetics Prof. Yong H. Won Spring 2010 3 Lesson 11 Poisson’s and Laplace’s Equations EE204B Electromagnetics Prof. Yong H. Won Spring 2010 4 Lesson 11 Poisson’s and Laplace’s Equations Poisson’s Equation in inhomogeneous medium EE204B Electromagnetics Prof. Yong H. Won Spring 2010 5 Lesson 11 Poisson’s and Laplace’s Equations In Cartesian Coordinate System ∂V r ∂V r ∂V r ax + ay + az ∂x ∂y ∂z ∂ 2V ∂ 2V ∂ 2V ∇ 2V = ∇ ⋅ (∇V ) = 2 + 2 + 2 ∂x ∂y ∂z ∇V = In Cylindrical Coordinate System 1∂ ∂V 1 ∂ 2V ∂ 2V ∇V= (ρ )+ 2 +2 2 ∂z ρ ∂ρ ∂ρ ρ ∂φ 2 In Spherical Coordinate System 1 ∂ 2 ∂V 1 ∂ ∂V 1 ∂ 2V ∇ 2V = 2 (r )+ 2 (sin θ )+ 2 2 r ∂r ∂r r sin θ ∂θ ∂θ r sin θ ∂φ 2 EE204B Electromagnetics Prof. Yong H. Won Spring 2010 6 Lesson 11 Uniqueness Theorem Consider the problem introduced at the beginning. th th z It can be described by the following z=d ∇ V =0 0≤ z≤d V =0 z = 0 ⎫ Boundary ⎬ Conditions z = d ⎭ (A & B are 2 V = 1V 1V z=0 r E =? boundaries boundaries of 0 ≤ z ≤ d) What is in the box is so called as a boundary value problem. The solution is unique! solution is unique! The solution to Laplace’s equation or Poisson’s equation) subject to certain boundary conditions is unique! subject to certain boundary conditions is unique! EE204B Electromagnetics Prof. Yong H. Won Spring 2010 7 Lesson 11 Solving Poisson’s or Laplace ’s Eq . Steps for solving either Poisson’s or Laplace’s equation analytically: 1. Choose the most appropriate representation of Laplacian based on any symmetry. 2. Perform the integration of the differential equation to obtain the most general solution for the potential. 3. Let this general solution to satisfy the boundary conditions to find constants of integration. diti fi Addi Additionally, numerical methods, such as Finite Difference Method, Finite Fi Diff Fi Element Method, Method of Moments, are developed to solve Boundary Value problems in situations where analytical solution is hard to find. EE204B Electromagnetics Prof. Yong H. Won Spring 2010 8 Lesson 11 Solving Poisson’s or Laplace ’s Eq . Example 11-1 z z=d z=0 1V Boundary value problem ∇2 V = 0 V=0 V=1V 0≤ z ≤d z=0 z=d Solution If ignoring the fringing field (valid when the plate width is much larger than the distance d), V can be regarded as a function of z only. EE204B Electromagnetics Prof. Yong H. Won Spring 2010 9 Lesson 11 Solving Poisson’s or Laplace ’s Eq . ∂ 2V ∇ V = 2 =0 ∂z 2 ⇒ V = Az + B To determine A and B, use boundary conditions determine and use boundary conditions z = 0, V = 0 ⇒ o=B − − − (1) z = d, V =1 ⇒ 1 = Ad + B − − − (2) From (1) and (2) 1 ⎧ ⎪A = ⇒⎨ d ⎪ ⎩B=0 1 ∴V = z d EE204B Electromagnetics r ∂V r 1r E = −∇V = − az = − az ∂z d Prof. Yong H. Won Spring 2010 E 10 Lesson 11 Solving Poisson’s or Laplace ’s Eq . Question What is the surface charge density ρs on the two plates? On the upper plate, th Assuming region is inside the plate and r an On the lower plate the lower plate r an EE204B Electromagnetics region is in air an is normal to the interface pointing to . r rr From D2n- D1n = ρs an ⋅ ( D2 − D1 ) = ρ s r rr r 1r Q D1 = 0 ⇒ ρ s = an ⋅ D2 = −az ⋅ (− azε 0 ) d ε0 ∴ ρs = d r r r From an ⋅ ( D2 − D1 ) = ρ s (Remember an is from 1 to 2) r rr r 1r Q D1 = 0 ⇒ ρ s = an ⋅ D2 = az ⋅ (− azε 0 ) d ε0 ∴ ρs = − d Prof. Yong H. Won Spring 2010 11 Lesson 11 Solving Poisson’s or Laplace ’s Eq . Example 11-2 An infinitely long coaxial cylindrical structure having an inner conductor of radius a, and an outer conductor of radius b. The inner Th conductor is attached to V0; the outer one is grounded. a V0 b What is E in between a and b? EE204B Electromagnetics Prof. Yong H. Won Spring 2010 12 Lesson 11 Solving Poisson’s or Laplace ’s Eq . Solution ∇ 2V = 0 a≤ ρ ≤b 1∂ ∂V 1 ∂ 2V ∂ 2V ∇V= (ρ )+ 2 + 2 =0 2 ρ ∂ρ ∂ρ ρ ∂φ ∂z 2 The structure has a cylindrical symmetry ∴ ∂V = 0, ∂φ i.e. V does not change with φ The structure has no variation along The structure has no variation along z ∂V =0 ∂z 1∂ ∂V ∂V (ρ )=0 ⇒ ρ ∴ = C (constant) ρ ∂ρ ∂ρ ∂ρ C ∴V = ∫ d ρ = C ln ρ + D ∴ ρ EE204B Electromagnetics Prof. Yong H. Won Spring 2010 13 Lesson 11 Solving Poisson’s or Laplace ’s Eq . To determine C and D, use boundary conditions, ρ = a, ρ = b, V = V0 V =0 Substitute them into the general solution th th C ln a + D = V0 ⎫ ⎬ C ln b + D = 0 ⎭ V= V0 ln(a / b) V0 D=− ln b ln(a / b) C= V0 V0 ln ρ − ln b ln(a / b) ln(a / b) r ∴ E = −∇V = − EE204B Electromagnetics a≤ ρ ≤b V0 1r aρ ln(a / b) ρ Prof. Yong H. Won Higher potential to lower potential, Perpendicular to the conducting surface Spring 2010 14 Lesson 11 Solving Poisson’s or Laplace ’s Eq . Example 11-3 Calculate the potential variation between two infinite parallel metal metal plates in a vacuum. x=0 x = x0 We assume no resistance, therefore, no variation in y and z directions. Since no charges exist between plates, we need to solve the Laplace’s equation: d 2V =0 2 dx Boundary conditions: V = V0 at x = 0; V = 0 at x = x0 EE204B Electromagnetics Prof. Yong H. Won Spring 2010 15 Lesson 11 Solving Poisson’s or Laplace ’s Eq . The solution of the Laplace’s equation will be in the form: V = C1 x + C2 Here C1 and C2 are integration constant that can be found from boundary cond. ⎧V0 = C1 ⋅ 0 + C2 ⎧C2 = V0 ⇒⎨ ⎨ ⎩0 = C1 ⋅ x0 + C2 ⎩C1 = − V0 x0 Finally, the solution: The electric field is: EE204B Electromagnetics ⎛ x⎞ V = V0 ⎜1 − ⎟ ⎝ x0 ⎠ V0 dV E=− ux = ux dx x0 Prof. Yong H. Won Spring 2010 16 Lesson 11 Solving Poisson’s or Laplace ’s Eq . Let’s assume that there is a charge with ρv uniformly distributed between plates if di ρ d 2V Poisson’s equation: =− v dx 2 ε0 ρv x 2 + C3 x + C4 The solution will be in form: V = − ε0 2 from the boundary conditions: ⎧ ρv 02 ⎧C4 = V0 ⎪V0 = − ε 2 + C3 ⋅ 0 + C4 ⎪ ⎪ 0 ⇒⎨ ρv x0 2 ⎞ 1⎛ ⎨ 2 ⎪0 = − ρv x0 + C x + C ⎪C3 = − x ⎜ V0 − ε 2 ⎟ 30 4 0⎝ 0 ⎠ ⎩ ⎪ ε0 2 ⎩ ⎛ ⎞⎛ ρv x⎞ V = V0 ⎜1 + x0 x ⎟⎜1 − ⎟ ⎝ 2ε 0V0 ⎠⎝ x0 ⎠ EE204B Electromagnetics V0 ⎛ ρ v x0 2 ⎛ x ⎞⎞ Ε = ⎜1 − ⎜ 2ε V ⎜1 − 2 x ⎟ ⎟ u x ⎟ x0 ⎝ 0 0⎝ 0 ⎠⎠ Prof. Yong H. Won Spring 2010 17 Lesson 11 Solving Poisson’s or Laplace ’s Eq . EE204B Electromagnetics Prof. Yong H. Won Spring 2010 18 Lesson 11 Solving Poisson’s or Laplace ’s Eq . Example 11-4 11 z d a ε2 ++++++++++++++++++++ --------------------- ε1 ρs 0 Both plates are grounded. In between there are two media characterized by ε1 and ε2, respectively. There exists surface charges on the dielectric interface, the density of which is on the dielectric interface, the density of which is ρs (C/m2). What is E between two plates? (Note: this example is slightly different from Text Example 6.2. ε1 and ε2 this example is slightly different from Text Example are exchanged. We used z-axis instead of x-axis.) EE204B Electromagnetics Prof. Yong H. Won Spring 2010 19 Lesson 11 Solving Poisson’s or Laplace ’s Eq . Solution The problem is inhomogeneous by itself, i.e., two different media are involved. However, it can be treated as two homogeneous regions: 0 ≤ z ≤ a and a ≤ z ≤ d In each region, V satisfies ∇ 2V = 0 Si Since ρv in each region is 0. So we will solve Lapalace’s equation ∇ 2V = 0 in each region, then combine the solutions. For 0 ≤ z ≤ a ∇ 2V = 0 ∂V ∂V ∂V + 2 + 2 =0 2 ∂x ∂y ∂z 2 2 2 EE204B Electromagnetics The structure is infinitely long along x, y V has no x-, y-dependence (i.e., V does not change with x and y.) Prof. Yong H. Won Spring 2010 20 Lesson 11 Solving Poisson’s or Laplace ’s Eq . ∂ 2V ∂V ∴ 2 =0 ⇒ = A (constant) ∂z ∂z ∴V = ∫ Adz = Az + B ( B is another constant) This is V for the region 0 ≤ z ≤ a, Let us denote it by V1. ∴V1 = Az + B For a ≤ z ≤ d ∇ V =0 2 ⇒ ∂ 2V =0 2 ∂z ⇒ V = Cz + D Denoting it by V2 V2 = Cz + D EE204B Electromagnetics Prof. Yong H. Won Spring 2010 21 Lesson 11 Solving Poisson’s or Laplace ’s Eq . To determine A,B,C and D, use boundary conditions, z = a, V1 = V2 Aa + B = Ca + D r r r D2 = ε 2 E2 = ε 2 (−∇V2 ) = ε 2 (−Caz ) r r r D1 = ε1 E1 = ε1 (−∇V1 ) = ε1 (− Aaz ) r rr r r r From an ⋅ ( D2 − D1 ) = ρ s ⇒ az ⋅ (−Cε 2 az + Aε1a ) = ρ s ⇒ − Cε 2 + Aε1 = ρ s Solve A, B, C , D from the above 4 box equations a d −a C = − ρs A = ρs (ε 2 − ε1 )a + ε1d (ε 2 − ε1 )a + ε1d B=0 EE204B Electromagnetics D = ( A − C )a = ρ s Prof. Yong H. Won d (ε 2 − ε1 )a + ε1d Spring 2010 22 Lesson 11 Solving Poisson’s or Laplace ’s Eq . d −a V1 = ρ s z (ε 2 − ε1 )a + ε1d 0≤ z≤a a d V2 = − ρ s z + ρs (ε 2 − ε1 )a + ε1d (ε 2 − ε1 )a + ε1d r d −a ⎧r ⎪ E = −∇V = − ρ s (ε − ε )a + ε d az ⎪ 2 1 1 ∴⎨ r r a ⎪ E = −∇V = − ρ az s ⎪ (ε 2 − ε1 )a + ε1d ⎩ a≤z≤d 0≤ z≤a a≤z≤d d a 0 EE204B Electromagnetics Prof. Yong H. Won Spring 2010 23 Lesson 11 Solving Poisson’s or Laplace ’s Eq . Example 11-5 V0 a b Two concentric spherical shells of radius a and b. The inner shell is attached to V0. The outer shell is grounded. What is E in region a ≤ r ≤ b ? EE204B Electromagnetics Prof. Yong H. Won Spring 2010 24 Lesson 11 Solving Poisson’s or Laplace ’s Eq . Solution ∇ 2V = 0 a≤r ≤b 1 ∂ 2 ∂V 1 ∂ ∂V 1 ∂ 2V =0 (r )+ 2 (sin θ )+ 2 2 2 2 ∂r ∂θ r ∂r r sin θ ∂θ r sin θ ∂φ Because of spherical symmetry ∂V ∂V = 0, =0 ∂θ ∂φ 1 ∂ 2 ∂V 2 ∂V ∴2 (r )=0 ⇒ r = C (constant) r ∂r ∂r ∂r C C ∴V = ∫ 2 dr = − + D r r EE204B Electromagnetics Prof. Yong H. Won Spring 2010 25 Lesson 11 Solving Poisson’s or Laplace ’s Eq . To determine C and D, use boundary conditions, r = a, V = V0 r = b, V =0 ab ⎧ ⎪ C = V0 a − b ⎪ ⇒⎨ ⎪D = V a 0 ⎪ a −b ⎩ ab 1 a ∴V = −V0 + V0 a≤r ≤b a −b r a −b r ab 1 r ∂V r ∴ E = −∇V = − ar = −V0 a 2r ∂r a −b r EE204B Electromagnetics Prof. Yong H. Won r E Spring 2010 V0 26 ...
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