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(b) From symmetry, we see in this case that the net field component along the
y
axis is
zero; the net field component along the
x
axis points rightward. With
θ
= 60
°
,
net,
2
0
cos
2
4
x
Q
E
a
πε
=
.
Since cos(60
°
) = 1/2, we can write this as
E
net
=
kQ/a
2
(using the notation of Eq. 215).
Thus,
E
net
≈
27 N/C.
65. (a) From symmetry, we see the net field component along the
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Unformatted text preview: x axis is zero; the net field component along the y axis points upward. With θ = 60 ° , net, 2 sin 2 4 y Q E a = . Since sin(60 ° ) = 3 /2 , we can write this as E net = kQ 3 /a 2 (using the notation of the constant k defined in Eq. 215). Numerically, this gives roughly 47 N/C....
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This note was uploaded on 07/10/2010 for the course PHYS 321 taught by Professor Wb during the Fall '10 term at The Petroleum Institute.
 Fall '10
 WB

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