mcconnell (kam2342) – oldhomework 05 – Turner – (60230)
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001
10.0 points
An electron, starting from rest, is accelerated
by a uniform electric field of 80000 N
/
C that
extends over a distance of 6 cm.
The charge and mass of an electron are
1
.
6
×
10
−
19
C and 9
.
11
×
10
−
31
kg respectively.
Find the speed of the electron after it leaves
the region of uniform electric field.
Correct answer: 4
.
10617
×
10
7
m
/
s.
Explanation:
Let
e
= 1
.
6
×
10
−
19
C
,
m
e
= 9
.
11
×
10
−
31
kg
,
E
= 80000 N
/
C
,
and
Δ
x
= 6 cm = 0
.
06 m
.
Because of the constant acceleration,
v
2
=
v
2
0
+ 2
a
Δ
x .
Since
v
0
= 0 and
a
=
F
net
m
e
=
e E
m
e
,
v
=
radicalbigg
2
e E
Δ
x
m
e
=
radicalBigg
2 (1
.
6
×
10
−
19
C) (80000 N
/
C)
9
.
11
×
10
−
31
kg
×
√
0
.
06 m
=
4
.
10617
×
10
7
m
/
s
.
002
(part 1 of 3) 10.0 points
An electron traveling at 4
×
10
6
m
/
s enters a
0
.
08 m region with a uniform electric field of
287 N
/
C
,
as in the figure.
The
mass
of
an
electron
is 9
.
10939
×
10
−
31
kg and the charge on an
electron is 1
.
60218
×
10
−
19
C
.









0
.
08 m
+
+
+
+
+
+
+
+
+
ˆ
ı
ˆ
4
×
10
6
m
/
s
Find the magnitude of the acceleration of
the electron while in the electric field.
Correct answer: 5
.
04781
×
10
13
m
/
s
2
.
Explanation:
Let :
q
e
= 1
.
60218
×
10
−
19
C
,
m
e
= 9
.
10939
×
10
−
31
kg
,
and
E
= 287 N
/
C
.
F
=
m a
=
q E ,
so
vectora
=

q
e
E
m
e
ˆ
=

(1
.
60218
×
10
−
19
C)(287 N
/
C)
9
.
10939
×
10
−
31
kg
ˆ
=

(5
.
04781
×
10
13
m
/
s
2
)ˆ
,
and the magnitude of the acceleration of the
electron is
5
.
04781
×
10
13
m
/
s
2
.
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 Spring '10
 Turner
 Electric charge, McConnell, vxf, Aconical surf ace

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