brown (twb493) – Ch14h2 – chiu – (56565)
1
This
printout
should
have
25
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
10.0points
A particle with charge +7 nC (a nanocoulomb
is 1
×
10

9
C) is located at the origin. At a
point
P
given by
vectorr
=
(
0
.
7 m
,
0
,
0
)
,
the electric field will be of the form
vector
E
=
(
E
x
,
0
,
0
)
.
Find
E
x
.
Correct answer: 128
.
394 N
/
C.
Explanation:
The general equation for the electric field
at
P
is
vector
E
=
1
4
πǫ
0
q

vectorr

2
ˆ
r.
The
x
component would be given by
E
x
=
1
4
πǫ
0
q

vectorr

2
ˆ
r
x
.
We have the various quantities that we
need:
q
= 7 nC = 7
×
10

9
C

vectorr

= 0
.
7 m
ˆ
r
x
= 1
.
Plugging in, we obtain
E
x
=
1
4
πǫ
0
7
×
10

9
C
(0
.
7 m)
2
= 128
.
394 N
/
C
.
002(part1of2)10.0points
You want to create an electric field
vector
E
=
(
0
,
3979 N
/
C
,
0
)
at the origin.
Find the
y
coordinate where
you would need to place a proton, in order to
create this field at the origin.
Correct answer:
−
6
.
01165
×
10

7
m.
Explanation:
The equation for the electric field, in gen
eral, is
vector
E
=
1
4
πǫ
0
q

vectorr

2
ˆ
r.
In this case,
q
= 1
.
6
×
10

19
C for a proton,
and
ˆ
r
=
(
0
,
1
,
0
)
,
since the field points in the positive
y
direc
tion.

vectorr

is the unknown we want to solve for.
We can write
E
y
= 3979 N
/
C =
1
4
πǫ
0
1
.
6
×
10

19
C

vectorr

2
⇒ 
vectorr

2
=
1
4
πǫ
0
1
.
6
×
10

19
C
3979 N
/
C
⇒ 
vectorr

=
±
radicalBigg
1
4
πǫ
0
1
.
6
×
10

19
C
3979 N
/
C
=
±
6
.
01165
×
10

7
m
.
We choose the minus sign to place the pro
ton below the origin; this way the field will
point upward like we want it to.
003(part2of2)10.0points
Instead of a proton, where would you place
an electron to produce the same field at the
origin?
Correct answer: 6
.
01165
×
10

7
m.
Explanation:
This is simple.
The proton and electron
carry the same magnitude of charge, but the
electron’s charge is negative. The calculation
from above will be exactly the same, but we
choose a positive sign at the end to place the
electron
above
the origin. This way the field
will still point upward as we want it to.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
brown (twb493) – Ch14h2 – chiu – (56565)
2
004(part1of6)10.0points
A
π

(“piminus”) particle, which has charge
−
e
, is at location
vectorr
π
=
(
8
×
10

9
,
−
3
×
10

9
,
−
5
×
10

9
)
m
.
You will find the components of the electric
field at another point
P
given by
vectorr
P
=
(−
7
×
10

9
,
7
×
10

9
,
4
×
10

9
)
m
due to the
π

particle. Begin by finding the
x
component of the field, which we will call
E
x
(
P
). Answer in N/C.
Correct answer: 2
.
63672
×
10
6
N
/
C.
Explanation:
Let’s write the general expression for the
electric field at
P
due to the
π

as
vector
E
Pπ
=
1
4
πǫ
0
Q
π


vectorr
Pπ

2
ˆ
r
Pπ
.
Therefore in order to determine the com
ponents of the field, we need to find the unit
vector ˆ
r
Pπ
pointing from the location of the
π

to the point
P
and the distance between
those two locations,
vextendsingle
vextendsingle
vextendsingle
vectorr
Pπ
vextendsingle
vextendsingle
vextendsingle
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 Turner
 Physics, Electron, Correct Answer, Electric charge, N/C

Click to edit the document details