201 Lab - i 1 .0 k0j 911 Ryan Johnson 2/11/09 9 0 Part I: 9...

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Unformatted text preview: i 1 .0 k0j 911 Ryan Johnson 2/11/09 9 0 Part I: 9 9 Q 11Lab 201: Electric Field by Point Charges 1 .0 Q 11 2 .0 2 . r ( x,y ) r ( x,y ) 2 Q 1 2 . 2x .i) (0). ( r1 (Py ) )(1i)(( x,(0j) Pxy) (x) r2(yj)) ) .jk 2 ..rr1 x,y )). E((,,y) k. ( (Q,y P (1 1 i x. 1y x, E ( x,y ) 1 2 2 i r(,y P,y Pr(,y P,y P 1x) ( ) 12x) ( ) 2 x x E,) Ex) Ex) (x 1(, 2(, yyy x 92,91.2 Ex1)1 E,y.i 1.1 9Ex) E,y.j ..1 . . x ,y0 9(x) 0 y(,y ( ) 91 Q . Q x k1 0 1( 2 1 j 0 0 1 1 0 Q 1 r ( x,y ) 1 Q 2 . 1 2x P (0)x,()1). ( r1 (Py ) )(0i)( x,(1j) Pxy) (x) r2(yj)) ) 1 E.i( y .jk 2 ..r1 y ). E((,,y) k. ( ( x.,y i x, 2 1 0 4 2 x 0 2 4 E ( x, 0) x 0 2 2 . r ( x,y ) r ( x,y ) 2 Part II: r(,y P,y Pr(,y P,y P 1x) ( ) 12x) ( ) 2 x x E,) Ex) Ex) (x 1(, 2(, yyy x 2, 1.2 Ex) E,y.i Ex) E,y.j 1 91 ( . . . x ,y (x) y(,y ( ) x 1 0 E ( x, 0) x 0 1 0 4 2 0 x 2 4 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. 4) Yes the electric field is equal to 0 at the origin, it would be expected to equal 0 because the charges are equally spaced from the origin so the opposite directions will cancel out. If a positive test charged were placed on the origin it would be in static equilibrium because the sum of the forces would be equal to 0. 5) a) According to the graph at the point (0,0) the magnitude of the electric field is still 0. b) Because the two charges are opposite, the electric field created between the two points should be high, while outside wither point the electric field should be zero when the charges cancel out. However, the electric field between the two charges is expected to be greater than zero because each charge is creating an electric field in the same direction between them. 1 0 Ryan Johnson 2/11/09 c) A positive test charge placed at the origin would not be in static equilibrium in this situation, because on positive charge at x = -1 would repel the test charge while the negative charge at x = 1 would attract the test charge to the right along the x axis. We also repeated this part of the experiment for part II: This graph actually shows the electric field on the x axis from -2.1 to +2.1, but since the charges are now on the y axis they do not create a field along the x axis, just intersecting it. ...
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This note was uploaded on 03/11/2010 for the course PHYS 121 taught by Professor Opyrchal during the Spring '08 term at NJIT.

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