16. The net field components along the
x
and
y
axes are
E
net
x
=
q
1
4
πε
o
R
2
–
q
2
cos
θ
4
πε
o
R
2
,
E
net
y
= –
q
2
sin
θ
4
πε
o
R
2
.
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
E
2
=
q
1
2
+
q
2
2
−
2
q
1
q
2
cos
θ
16
π
2
ε
o
2
R
4
.
With
R =
0.500 m,
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 Fall '10
 Rebello,NobelS
 Continuous function, Inverse function, Inverse trigonometric functions, Square number

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