marout (ssm979) – oldhomework 05 – Turner – (56725)
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001
10.0 points
An electron begins at rest, and then is accel
erated by a uniform electric field of 800 N
/
C
that extends over a distance of 6 cm.
Find the speed of the electron after it leaves
the region of uniform electric field.
The elementary charge is 1
.
6
×
10
−
19
C and
the mass of the electron is 9
.
11
×
10
−
31
kg.
Correct answer: 4
.
10617
×
10
6
m
/
s.
Explanation:
Let :
e
= 1
.
6
×
10
−
19
C
,
m
e
= 9
.
11
×
10
−
31
kg
,
E
= 800 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) (800 N
/
C)
9
.
11
×
10
−
31
kg
×
√
0
.
06 m
=
4
.
10617
×
10
6
m
/
s
.
002
10.0 points
A particle of mass 0
.
000109 g and charge
38 mC moves in a region of space where the
electric field is uniform and is 3
.
9 N
/
C in the
x direction and zero in the y and z direction.
If the initial velocity of the particle is given
by
v
y
= 1
.
1
×
10
5
m
/
s,
v
x
=
v
z
= 0, what is
the speed of the particle at 0
.
2 s?
Correct answer: 2
.
93333
×
10
5
m
/
s.
Explanation:
Let :
m
= 0
.
000109 g = 1
.
09
×
10
−
7
kg
,
E
x
= 3
.
9 N
/
C
,
E
y
=
E
z
= 0
,
v
y
= 1
.
1
×
10
5
m
/
s
,
v
x
=
v
z
= 0
,
and
t
= 0
.
2 s
.
According to Newton’s second law and the
definition of an electric field,
vector
F
=
mvectora
=
q
vector
E .
Since the electric field has only an
x
compo
nent, the particle accelerates only in the
x
direction
a
x
=
q E
x
m
.
To determine the
x
component of the final
velocity,
v
xf
, use the kinematic relation
v
xf
=
v
xi
+
a
(
t
f

t
i
) =
a t
f
.
Since
t
i
= 0 and
v
xi
= 0
,
v
xf
=
q E
x
t
f
m
v
xf
=
(0
.
038 C) (3
.
9 N
/
C)(0
.
2 s)
(1
.
09
×
10
−
7
kg)
= 2
.
71927
×
10
5
m
/
s
.
No external force acts on the particle in the
y
direction so
v
yi
=
v
yf
= 1
.
1
×
10
5
m
/
s. Hence
the final speed is given by
v
f
=
radicalBig
v
2
yf
+
v
2
xf
=
bracketleftbigg
(
1
.
1
×
10
5
m
/
s
)
2
+
(
2
.
71927
×
10
5
m
/
s
)
2
bracketrightbigg
1
/
2
=
2
.
93333
×
10
5
m
/
s
.
Note:
This is analogous to a particle in a
gravitational field with the coordinates ro
tated clockwise by
π
2
(90
◦
).
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marout (ssm979) – oldhomework 05 – Turner – (56725)
2
003
(part 1 of 2) 10.0 points
A cone with base radius
r
and height
h
is
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 Fall '10
 TURNER,J
 Electric charge, Aconical surf ace

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