saldana (avs387) – homework 08 – Turner – (56705)
1
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12
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001
(part 1 of 2) 10.0 points
A proton is accelerated through a potential
difference of 4
.
4
×
10
6
V.
a) How much kinetic energy has the proton
acquired?
Correct answer: 7
.
04
×
10
−
13
J.
Explanation:
Let :
Δ
V
= 4
.
4
×
10
6
V
and
q
= 1
.
60
×
10
−
19
C
.
Δ
K
= Δ
U
=
q
Δ
V
= (1
.
60
×
10
−
19
C) (4
.
4
×
10
6
V)
=
7
.
04
×
10
−
13
J
.
002
(part 2 of 2) 10.0 points
b) If the proton started at rest, how fast is it
moving?
Correct answer: 2
.
90104
×
10
7
m
/
s.
Explanation:
Let :
m
= 1
.
673
×
10
−
27
kg
.
Since
K
i
= 0 J
,
Δ
K
=
K
f
=
1
2
m v
2
f
v
f
=
radicalbigg
2
K
f
m
=
radicalBigg
2 (7
.
04
×
10
−
13
J)
1
.
673
×
10
−
27
kg
=
2
.
90104
×
10
7
m
/
s
.
003
10.0 points
Points A (1 m, 4 m) and B (3 m, 7 m) are in a
region where the electric field is uniform and
given by
vector
E
=
E
x
ˆ
ı
+
E
y
ˆ
, where
E
x
= 4 N
/
C
and
E
y
= 3 N
/
C.
What is the potential difference
V
A

V
B
?
Correct answer: 17 V.
Explanation:
Let :
E
x
= 4 N
/
C
,
E
y
= 3 N
/
C
,
(
x
A
, y
A
) = (1 m
,
4 m)
,
and
(
x
B
, y
B
) = (3 m
,
7 m)
.
We know
V
(
A
)

V
(
B
) =

integraldisplay
A
B
vector
E
·
dvectors
=
integraldisplay
B
A
vector
E
·
dvectors
For a uniform electric field
vector
E
=
E
x
ˆ
ı
+
E
y
ˆ
.
Now consider the term
E
x
ˆ
ı
·
dvectors
in the inte
grand.
E
x
is just a constant and ˆ
ı
·
dvectors
may be
interpreted as the projection of
dvectors
onto
x
, so
that
E
x
ˆ
ı
·
dvectors
=
E
x
dx .
Likewise
E
y
ˆ
·
dvectors
=
E
y
dy .
Or more simply,
dvectors
=
dx
ˆ
ı
+
dy
ˆ
dotting it
with
E
x
ˆ
ı
+
E
y
ˆ
gives the same result as above.
Therefore
V
A

V
B
=
E
x
integraldisplay
x
B
x
A
dx
+
E
y
integraldisplay
y
B
y
A
dy
= (4 N
/
C) (3 m

1 m)
+ (3 N
/
C) (7 m

4 m)
=
17 V
.
Note that the potential difference is inde
pendent of the path taken from A to B.
004
10.0 points
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saldana (avs387) – homework 08 – Turner – (56705)
2
Consider a circular arc of constant linear
charge density
λ
as shown below.
x
y
3
7
π
+
+
+
+
+
+
+
+
+
+
+
O
r
What is the potential
V
O
at the origin
O
due to this arc?
1.
V
O
=
5
36
λ
ǫ
0
2.
V
O
=
3
28
λ
ǫ
0
correct
3.
V
O
=
5
24
λ
ǫ
0
4.
V
O
=
1
7
λ
ǫ
0
5.
V
O
=
3
14
λ
ǫ
0
6.
V
O
=
5
32
λ
ǫ
0
7.
V
O
=
5
28
λ
ǫ
0
8.
V
O
= 0
9.
V
O
=
1
5
λ
ǫ
0
10.
V
O
=
3
22
λ
ǫ
0
Explanation:
The potential at a point due to a continuous
charge distribution can be found using
V
=
k
e
integraldisplay
dq
r
.
In this case, with linear charge density
λ ,
dq
=
λ ds
=
λ r dθ ,
so
V
=
k
e
integraldisplay
3
7
π
0
λ dθ
=
1
4
π ǫ
0
integraldisplay
3
7
π
0
λ dθ
=
λ
4
π ǫ
0
θ
vextendsingle
vextendsingle
vextendsingle
vextendsingle
3
7
π
0
=
λ
4
π ǫ
0
parenleftbigg
3
7
π

0
parenrightbigg
=
3
28
λ
ǫ
0
.
005
10.0 points
A dipole field pattern is shown in the figure.
Consider various relationships between the
electric potential at different points given in
the figure.
Z
N
D
J
K
+

Notice:
Five potential relationships are
given below.
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 Spring '10
 Turner
 Charge, Electrostatics, Work, Electric charge, Saldaña

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