TST2-351-F08-Solution

TST2-351-F08-Solution - No Name on this page - only on the...

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Unformatted text preview: No Name on this page - only on the back. Thanks! Physics 351 Elec. & Magnetism Test 2 Prof. Asim Gangopadhyaya As always, to receive full credit you must show your work in a neat, organized and a legible form. Some Relevant Formulae are given below: V sph ( r,, ) = X =0 A r + B r +1 P (cos ) V cyl ( , ) = ( a + b log )( c + d ) + X n =1 ( a n n + b n - n ) ( c n cos( n ) + d n sin( n )) ~ E 2- ~ E 1 n 1 2 = ; ~ E 2- ~ E 1 n 1 2 = 0 ; cos( a + b ) = cos a cos b- sin a sin b , P ( x ) = 1 , P 1 ( x ) = x , P 2 ( x ) = 1 2 (3 x 2- 1) , P 3 ( x ) = 1 2 (5 x 3- 3 x ) ; Z 1- 1 dx P m ( x ) P n ( x ) = 2 2 m + 1 mn (1 + x ) = 1 + x + ( - 1) 2! x 2 + ( - 1)( - 2) 3! x 3 + ( - 1)( - 2)( - 3) 4! x 4 + ; converges for | x | < 1 . ln(1 + x ) = x- x 2 2 + x 3 3 + + (- 1) n +1 x n n + 1. (10 points) Electrostatic field inside a conductor is zero. Prove that Electrostatic field just outside a conductor is always perpendicular to the surface. Solution: ~ E 2- ~ E 1 n 1 2 = 0 implies that tangential component of electric field is continuous across any interface. Since ~ E in = 0, its tangential component is zero as well. Which means that tangential component of ~ E out must be zero. Thus, only the normal component outside is potentiall non-zero.must be zero....
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TST2-351-F08-Solution - No Name on this page - only on the...

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