soto (rrs946) – Homework 4 – Hofmann – (56605)
1
This printout should have 20 questions.
Multiplechoice questions may continue on
the next column or page – Fnd all choices
be±ore answering.
001 (part 1 of 3) 10.0 points
A proton is released ±rom rest in a uni±orm
electric Feld o± magnitude 40000 V
/
m di
rected along the positive
x
axis. The proton
undergoes a displacement o± 0
.
2 m in the di
rection o± the electric Feld as shown in the
Fgure.
The mass o± a proton is 1
.
672623
×
10
−
27
kg.
+
+
+
+
+
+
+
+
+
+
+
+
+













v
A
= 0
+
0
.
2 m
v
0
40000 V
/
m
A
B
²ind the change in the electric potential i±
the proton moves ±rom the point
A
to
B
.
Correct answer:

8000 V.
Explanation:
Let :
q
p
= 1
.
60218
×
10
−
19
C
,
m
p
= 1
.
67262
×
10
−
27
kg
,
E
= 40000 V
/
m
,
and
x
= 0
.
2 m
.
²rom the equation
Δ
V
=

±
E
·
i
B
A
d±s
or in a noncalculus way ±or the straight path
AB
,
Δ
V
=

±
E
·
±
AB
we obtain
Δ
V
=

E x
=

(40000 V
/
m) (0
.
2 m)
=

8000 V
.
The negative result tells us that the electric
potential o± the proton decreases as it moves
±rom
A
to
B
.
002 (part 2 of 3) 10.0 points
²ind the change in potential energy o± the
proton ±or this displacement.
Correct answer:

1
.
28174
×
10
−
15
J.
Explanation:
The change in the potential energy o± the
proton is
Δ
U
=
q
p
Δ
V
= (1
.
60218
×
10
−
19
C) (

8000 V)
=

1
.
28174
×
10
−
15
J
.
The negative sign here means that the poten
tial energy o± the proton decreases as it moves
in the direction o± the electric Feld.
This
makes sense since, as the proton accelerates
in the direction o± the Feld, it gains kinetic
energy and at the same time (the Feld) loses
potential energy (the energy is conserved).
003 (part 3 of 3) 10.0 points
Apply the principle o± energy conservation to
Fnd the speed o± the proton a±ter it has moved
0
.
2 m, starting ±rom rest.
Correct answer: 1
.
23799
×
10
6
m
/
s.
Explanation:
Conservation o± energy in this case is
Δ
K
+ Δ
U
= 0
.
K
f

K
i
+ Δ
U
= 0
1
2
m
p
v
2
f
+ Δ
U
= 0
v
f
=
r

2 Δ
U
m
p
=
r

2 (

1
.
28174
×
10
−
15
J)
1
.
67262
×
10
−
27
kg
=
1
.
23799
×
10
6
m
/
s
.
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2
004 (part 1 of 2) 10.0 points
An electron moving parallel to the
x
axis has
an initial speed oF 5
×
10
6
m
/
s at the origin.
Its speed is reduced to 1
×
10
5
m
/
s at the point
x
P
, 5 cm away From the origin. The mass oF
the electron is 9
.
10939
×
10
−
31
kg and the
charge oF the electron is

1
.
60218
×
10
−
19
C.
Calculate the absolute value oF the poten
tial diference between this point and the ori
gin.
Correct answer: 71
.
042 V.
Explanation:
Let :
m
e
= 9
.
10939
×
10
−
31
kg
,
q
e
=

1
.
60218
×
10
−
19
C
,
v
i
= 5
×
10
6
m
/
s
,
v
f
= 1
×
10
5
m
/
s
,
and
d
= 5 cm
.
By conservation oF energy
(
K
+
U
)
f
= (
K
+
U
)
i
U
f

U
i
≡
Δ
U
=
K
i

K
f
,
Using the de±nition oF kinetic energy and solv
ing For Δ
U
yields
Δ
U
=
1
2
m
e
(
v
2
i

v
2
f
)
=
1
2
(9
.
10939
×
10
−
31
kg)
×
b
(
5
×
10
6
m
/
s
)
2

(1
×
10
5
m
/
s)
2
B
= 1
.
13822
×
10
−
17
J
,
²inally, the potential diference is
Δ
V
=
Δ
U
q
e
=
1
.
13822
×
10
−
17
J
(

1
.
60218
×
10
−
19
C)
=

71
.
042 V
.
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 Fall '11
 HOFFMAN
 Electrostatics, Conservation Of Energy, Energy, Potential Energy, Work, Electric charge, Soto

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