aee1_tutorials_solution

aee1_tutorials_solution - Electromagnetics Tutorial Prof....

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Unformatted text preview: Electromagnetics Tutorial Prof. S. Peik Tutorial 1 : Charges, Electrostatics Solution 1: Left and right each side Q s = λ 1 2 m 2 · 1 m = λ · m (1) Center piece: Q c = integraldisplay 1 − 1 λ x 2 dx = λ 1 3 ( 1 + 1 ) m = 2 λ 3 m (2) Total is Q = 8 3 λ m 2. left and right cancels out hence: vector r = and vector r ′ = x ˆ x + 1 ˆ y | vector r − vector r ′ | = √ x 2 + 1 Now vector E = 1 4 πε integraldisplay C λ ( vector r − vector r ′ ) | vector r − vector r ′ | 3 dl ′ (3) = 1 4 πε integraldisplay 1 − 1 λ x 2 ( − x ˆ x − 1 ˆ y ) √ 1 + x 2 3 dx (4) = λ 4 πε parenleftBigg integraldisplay 1 − 1 − x 3 ˆ x √ 1 + x 2 3 dx + integraldisplay 1 − 1 − 1 ˆ y ) √ 1 + x 2 3 dx parenrightBigg (5) = − λ 4 πε braceleftBigg 1 − 1 ˆ x + bracketleftbigg x √ 1 + x 2 bracketrightbigg 1 − 1 ˆ y bracerightBigg (6) = − λ 4 πε braceleftbigg 0 ˆ x + parenleftbigg 1 √ 2 − − 1 √ 2 parenrightbigg ˆ y bracerightbigg (7) = − λ 4 πε √ 2 ˆ y (8) 3. V ( vector r ) = 1 4 πε integraldisplay V ρ ( vector r ′ ) 1 | vector r − vector r ′ | dV ′ (9) = 1 4 πε integraldisplay 1 − 1 λ x 2 √ x 2 + 1 dx (10) = 1 4 πε bracketleftbigg 1 2 x radicalbig x 2 + 1 −...
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This note was uploaded on 05/04/2010 for the course EECS 320256 taught by Professor Peik during the Fall '09 term at Jacobs University Bremen.

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aee1_tutorials_solution - Electromagnetics Tutorial Prof....

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