Physics 21
Spring,
2011
Solution to HW3
2212
Figure 2234 shows an uneven arrangement of elec
trons (
e
) and protons (
p
) on a circular arc of radius
r
=
1
.
95cm, with angles
θ
1
= 31
.
0
◦
,
θ
2
= 47
.
0
◦
,
θ
3
= 39
.
0
◦
,
and
θ
4
= 22
.
0
◦
. What are the (a) magnitude and (b) di
rection (relative to the positive direction of the x axis) of
the net electric field produced at the center of the arc?
Let us redefine the given angles so that we are measuring
them from the positive x axis:
θ
1
= 31
.
0
◦
,
θ
2
= 78
.
0
◦
,
θ
3
= 119
.
0
◦
, and
θ
4
= 158
.
0
◦
. Now, we can easily convert
the positions of the particles from (
r,θ
) notation to (
x,y
)
notation.
The net electric field can be calculated as the
vector sum of the individual electric fields of each particle:
E
=
E
0
+
E
1
+
E
2
+
E
3
+
E
4
. Here, particle 0 is the electron
on the x axis. Each particle’s electric field can be calculated
using the vector equation
E
i
=
k
q
i
(
r
−
r
i
)

r
−
r
i

3
where
r
is the loca
tion of the point where we are calculating the electric field,
and
r
i
is the location of particle
i
.
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 Spring '08
 Hickman
 Physics, Electron, Fundamental physics concepts, Simple machine

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