This preview shows page 1. Sign up to view the full content.
52. Let
q
1
denote the charge at
y
=
d
and
q
2
denote the charge at
y
=
−
d
. The individual magnitudes
¯
¯
¯
~
E
1
¯
¯
¯
and
¯
¯
¯
~
E
2
¯
¯
¯
are fgured From Eq. 233, where the absolute value signs For
q
are unnecessary since these
charges are both positive. The distance From
q
1
to a point on the
x
axis is the same as the distance From
q
2
to a point on the
x
axis:
r
=
√
x
2
+
d
2
. By symmetry, the
y
component oF the net feld along the
x
axis is zero. The
x
component oF the net feld, evaluated at points on the positive
x
axis, is
E
x
=2
µ
1
4
πε
0
¶µ
q
x
2
+
d
2
x
√
x
2
+
d
2
¶
where the last Factor is cos
θ
=
x/r
with
θ
being the angle For each individual feld as measured From the
x
axis.
(a) IF we simpliFy the above expression, and plug in
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Charge

Click to edit the document details