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Unformatted text preview: i Ryan Johnson
2/11/09
9
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Q 11Lab 201: Electric Field by Point Charges
1 .0 Q 11
2 .0 Part I: 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 i 2 2 2
. r ( x,y )
r ( x,y )
2 2 2
. r ( x,y )
r ( x,y )
2 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
0 1 E ( x, 0)
x 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 0 2 4 x 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 09/16/2011 for the course PHYSICS 121 taught by Professor Fayngold during the Spring '10 term at NJIT.
 Spring '10
 Fayngold
 Physics, Charge

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