SampleTest2Solution

SampleTest2Solution - ± π √ m 2 n 2 ² This leads to...

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Physics 351 Electricity and Magnetism Test 2 Prof. Asim Gangopadhyaya A cubical cavity sits with one end at the origin. It has no charge inside it. The potentials on the planar surfaces of the cavity are given by: V (0 , y, z ) = V 0 , V ( π, y, z ) = 0, V ( x, 0 , z ) = 0, V ( x, π, z ) = 0, V ( x, y, 0) = 0, V ( x, y, π ) = 0. Determine potential function for a general point inside the cube. solution: General solution will be of the form V ( x, y, z ) = X m,n A mn sin( m · y ) sin( n · z ) sinh ± ( π - x ) · m 2 + n 2 ² . To determine the coefficients A mn , we use the last BC, i.e., V (0 , y, z ) = V 0 . V (0 , y, z ) = V 0 = X m,n A mn sin( m · y ) sin( n · z ) sinh
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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 ( π-x ) · √ m 2 + n 2 ´ m · n · π 2 · sinh ³ ( π ) · √ m 2 + n 2 ´ 1...
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