Kapoor (mk9499) – oldhomework 03 – Turner – (60230)
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001
(part 1 of 4) 10.0 points
A uniformly charged circular arc
AB
of radius
R
is shown in the figure. It covers a quarter
of a circle and it is located in the second
quadrant.
The total charge on the arc is
Q >
0.
x
y
+
+
+
+
+
+
+
+
+
Δ
θ
θ
R
x
y
I
II
III
IV
B
A
O
Δ
s
≡
R
Δ
θ
The direction of the electric field
E
due to
the charge distribution at the origin is
1.
along the negative
x
axis.
2.
in quadrant IV.
correct
3.
along the negative
y
axis.
4.
along the positive
y
axis.
5.
in quadrant I.
6.
in quadrant II.
7.
in quadrant III.
8.
along the positive
x
axis.
Explanation:
The electric field for a positive charge is
directed away from it. In this case, the electric
field generated by each Δ
q
is going to be in
quadrant IV, so the total electric field will be
in the same quadrant.
002
(part 2 of 4) 10.0 points
Determine Δ
E
x
, the
x
component of the elec
tric field vector at the origin
O
due to the
charge element Δ
q
located at an angle
θ
sub
tended by an angular interval Δ
θ
.
1.
Δ
E
x
=
k Q
R
Δ
θ
π
sin
θ
2.
Δ
E
x
=
k Q
R
2
2 Δ
θ
π
sin
θ
3.
Δ
E
x
=
k Q
R
2
2 Δ
θ
π
cos
θ
correct
4.
Δ
E
x
=
k Q
R
2
2 Δ
θ
cos
θ
5.
Δ
E
x
=
k Q
2
R
2 Δ
θ
π
cos
θ
6.
Δ
E
x
=
k Q
R
2 Δ
θ
π
cos
θ
7.
Δ
E
x
= 0
8.
Δ
E
x
=
k Q
R
2
Δ
θ
sin
θ
9.
Δ
E
x
=
k Q
2
R
Δ
θ
cos
θ
10.
Δ
E
x
=
k Q
2
R
2
2 Δ
θ
π
cos
θ
Explanation:
Δ
E
=
k
Δ
q
R
2
.
The charge Δ
q
per unit arclength Δ
s
is
λ
≡
Q
π
2
R
=
2
Q
π R
,
so
Δ
q
=
λ
Δ
s
=
λ R
Δ
θ
=
2
Q
π R
R
Δ
θ
=
2
Q
π
Δ
θ .
On the other hand, the vector Δ
vector
E
makes an
angle
θ
with the
x
axis, so
Δ
E
x
=
k Q
R
2
2 Δ
θ
π
cos
θ
.
003
(part 3 of 4) 10.0 points
Find
E
x
, the electric field at the origin due
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Kapoor (mk9499) – oldhomework 03 – Turner – (60230)
2
to the full arc length for the case where
Q
=
3
.
7
μ
C and
R
= 1
.
74 m
.
Correct answer: 6992
.
38 N
/
C.
Explanation:
Let :
Q
= 3
.
7
μ
C
and
R
= 1
.
74 m
.
E
x
can be found by integrating the contribu
tions of all the Δ
q
’s in the arc. We get
E
x
=
integraldisplay
π/
2
0
k Q
R
2
2
π
cos
θ dθ
=
2
k Q
π R
2
=
2
k
(3
.
7
μ
C)
π
(1
.
74 m)
2
=
6992
.
38 N
/
C
.
004
(part 4 of 4) 10.0 points
The charge on an electron is
e
.
If at
O
we have
E
x
, then the magnitude of
the force
bardbl
vector
F
e
bardbl
on an electron at this point is
1.
bardbl
vector
F
e
bardbl
=
E
x
2
e
.
2.
bardbl
vector
F
e
bardbl
= 4
e E
x
.
3.
bardbl
vector
F
e
bardbl
=
e E
x
.
4.
bardbl
vector
F
e
bardbl
=
2
E
x
3
e
.
5.
bardbl
vector
F
e
bardbl
=
3
e E
x
2
.
6.
bardbl
vector
F
e
bardbl
=
E
x
3
e
.
7.
bardbl
vector
F
e
bardbl
=
√
2
e E
x
.
correct
8.
bardbl
vector
F
e
bardbl
=
E
x
e
√
2
.
9.
bardbl
vector
F
e
bardbl
= 3
e E
x
.
10.
bardbl
vector
F
e
bardbl
= 2
e E
x
.
Explanation:
E
x
E
y
E
From the symmetry of the charge distribu
tion, it can be seen that
E
y
=
E
x
, so
E
=
radicalBig
E
2
x
+
E
2
y
=
√
2
E
x
.
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 Fall '08
 Turner
 Charge, Electrostatics, Work, Electric charge, Kapoor

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