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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 Spring '08
 Opyrchal
 Charge, Electric charge, 1J, 0j, 2 1 j, Ryan Johnson

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