marout (ssm979) – extra
credit 01 – turner – (56725)
1
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001
10.0 points
Given symmetrically placed rectangular insu
lators with uniFormly charged distributions oF
equal magnitude as shown in the fgure.
x
y
+ + + + + +
     
In this fgure, at the origin, the net feld
v
E
net
is
1.
along 45
◦
direction in the frst quadrant.
2.
aligned with the positive
y
axis.
3.
along 60
◦
direction in the frst quadrant.
4.
along 225
◦
direction in the frst quad
rant.
5.
aligned with the positive
x
axis.
6.
aligned with the negative
y
axis.
correct
7.
along 135
◦
direction in the frst quad
rant.
8.
zero and the direction is undefned.
9.
along 30
◦
direction in the frst quadrant.
10.
aligned with the negative
x
axis.
Explanation:
At the origin the felds From the bottom slab
points towards the negative
y
axis (towards
the negatively charged slab).
The feld From the right and leFt slabs have
equal magnitudes at the origin and are sym
metric about the
y
axis. The feld From the
right and leFt slabs point away From each pos
itively charged slab. ThereFore the
x
compo
nent oF the feld cancels at the origin. The
net feld From the right and leFt slabs points
towards the negative
y
axis (away From the
positively charged slab).
Hence, the sum oF the electric felds From all
three slabs is aligned with the negative
y
axis.
002
10.0 points
A line oF charge starts at
x
=
x
0
, where
x
0
is positive, and extends along the
x
axis to
positive infnity.
IF the linear charge den
sity is given by
λ
=
λ
0
x
0
/x
, where
λ
0
is a
constant, determine the electric feld at the
origin. (Here ˆ
ı
denotes the unit vector in the
positive
x
direction.)
1.
k λ
0
2
x
0
(

ˆ
ı
)
correct
2.
k λ
2
0
2
x
0
(ˆ
ı
)
3.
k λ
0
2
x
0
ı
)
4.
k λ
0
x
0
ı
)
5.
k λ
0
x
0
(

ˆ
ı
)
6.
k λ
0
2
x
2
0
ı
)
Explanation:
±irst we realize that we are dealing with a
continuous
distribution oF charge (as opposed
to point charges). We must divide the dis
tribution into small elements and integrate.
Using Coulomb’s law, the electric feld cre
ated by each small element with charge
dq
is
dE
=
k dq
x
2
where
dq
=
λdx
=
λ
0
x
0
x
dx.
Now we integrate over the entire distribution
(i.e. From
x
=
x
0
to
x
= +
∞
) and insert our
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credit 01 – turner – (56725)
2
dq
:

v
E

=
i
k dq
x
2
=
i
∞
x
0
k λ
0
x
0
dx
x
3
=

k λ
0
x
0
2
1
x
2
v
v
v
∞
x
0
=
k λ
0
2
x
0
.
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 Fall '08
 Turner
 Physics, Charge, Electric charge

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