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201 Lab - i 1.0 k0j 911 Ryan Johnson 9 0 Part I 9 9 Q 11Lab...

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Ryan Johnson 2/11/09 Lab 201: Electric Field by Point Charges Part I: . . . . . . . . . r1xy , ( )Pxy , ( )P 1 xy , ( )Pxy , ( )P 2 . . . . Exy , ( ) E 1xy , ( )E 2x ( 2 1 1 .9 , 2 .1 . y ) Exy , ( )i . E xy , ( )Exy , ( )j . E x x 0 , ( ) x 4 2 0 2 4 10 0 10 Part II: . . . . . . . . . . . . . E x x 0 , ( ) x 4 2 0 2 4 10 0 10
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Ryan Johnson 2/11/09 Questions: 1) In part I you must exclude the points x= -1 and x= 1 because these points do not exist and there are vertical asymptotes because these points cause r1 or r2 to be equal to 0 which causes E1 or E2 to be undefined, therefore nonexistent. 2) The equations above vary as distance cubed in the denominator because after the calculation it will be multiplied by r to make it r squared in the denominator. 3) (See graph from Part I on previous page) The electric field does = 0 at the origin because the system is in equilibrium, and it would be expected to be so. If a positive test charge was placed at the origin it would be in equilibrium because the electric field around it is 0, therefore all forces on one side will cancel out an equal force in the opposite direction.
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