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EEE340Lecture 171 The E-field and surface charge density are And (4.15)
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EEE340Lecture 172 Example P.4-23: Determine the potential and E-field between two wedge-shaped semi-infinite plates, 0< < Solution: The voltage is a function of only, I.e. V( ). The Laplace equation reduces to The boundary conditions are Therefore and VoVo x y
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EEE340Lecture 173 4-3: Uniqueness of Electrostatic Solution A solution to the Laplace’s (or Poisson’s) equation that satisfies the given boundary conditions is the (only) unique solution, regardless what method was employed. Proof: Textbook pp. 158-159. 4-4: Method of Image
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EEE340Lecture 174 Observations: At any point on the ground As R As V is an even function w.s.t. the y-axis, i.e., The E-field is ground. GND y x Q (0,d,0)
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EEE340Lecture 175 On the other hand, Conclusions: This problem is equivalent to a configuration of an image charge of –Q, and then remove the GND. The equivalence is only for the upper half plane. The lower half plane has zero V and zero. In other words, the image solution in the lower half plane must be dropped.
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EEE340Lecture 176 Example 4-3: Image method for a conducting corner Solution where d1 d2 Q -Q Q
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EEE340Lecture 177 In the source free region, Poisson’s equation reduces to the Laplace equation A standard way to solve the PDE is the method of separation of variables, by Fourier (4-81) §4-5 Boundary Value Problems in Cartesian Coordinates
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EEE340Lecture 178 Example 4-7 Two-dimensional Laplace’s Equations with boundary conditions : 1.Separation V(x, y) = X(x) Y(y). X”Y + XY” = 0 0 b a x y
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EEE340Lecture 179 Let us assume From 1).
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EEE340Lecture 1710 Boundary conditions:
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EEE340Lecture 1711 Boundary conditions: Hence, or
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EEE340Lecture 1712 2. Find unknown constant C n. 1.) Superposition (Linear Space) Let
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