16. The net field components along the
x
and
y
axes are
12
2
net,
net,
22
2
00
0
cos
sin
,.
44
4
xy
qq
q
EE
R
RR
θ
πε
=−
=
−
The magnitude is the square root of the sum of the componentssquared.
Setting the
magnitude equal to
E
= 2.00
×
10
5
N/C, squaring and simplifying, we obtain
2
11
1
2
0
2c
o
s
(4
)
q
q
E
R
+−
=
.
With
R =
0.500 m,
q
1
= 2.00
×
10
−
6
C and
q
2
= 6.00
×
10
−
6
C, we can solve this
expression for cos
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 Spring '10
 BB
 Physics, Continuous function, Inverse function, Inverse trigonometric functions, Square number

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