Unformatted text preview: ± ( π ) · √ m 2 + n 2 ² . This leads to A mn = 4 V ± cos( m · y )  π m ² ± cos( n · z )  π n ² π 2 · sinh ³ ( π ) · √ m 2 + n 2 ´ = 4 V (1(1) m ) (1(1) n ) m · n · π 2 · sinh ³ ( π ) · √ m 2 + n 2 ´ Thus the ﬁnal form of the solution is V ( x, y, z ) = X m,n 4 V (1(1) m ) (1(1) n ) sin( m · y ) sin( n · z ) sinh ³ ( πx ) · √ m 2 + n 2 ´ m · n · π 2 · sinh ³ ( π ) · √ m 2 + n 2 ´ 1...
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This note was uploaded on 01/28/2010 for the course PHYS 351 taught by Professor Gangopashyaya during the Spring '09 term at Loyola Chicago.
 Spring '09
 Gangopashyaya
 Charge, Magnetism

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