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fullwood (tf5349) – Homework 1 Electrostatics – Catala – (12054)
1
This printout should have 22 questions.
Multiplechoice questions may continue on
the next column or page – Fnd all choices
before answering.
001
10.0 points
±ind the magnitude of the electric Feld at a
point midway between two charges +17
.
2
×
10
−
9
C and +62
.
9
×
10
−
9
C separated by a
distance of 77.2 cm. The value of the Coulomb
constant is 8
.
99
×
10
9
N
·
m
2
/
C
2
.
Correct answer: 2757
.
41 N
/
C.
Explanation:
Let :
q
1
= 17
.
2
×
10
−
9
C
,
q
2
= 62
.
9
×
10
−
9
C
,
x
= 77
.
2 cm
,
and
k
C
= 8
.
99
×
10
9
N
·
m
2
/
C
2
.
r
1
=
r
2
=
x
2
=
77
.
2 cm
2
·
p
1 m
100 cm
P
= 0
.
386 m
The electric Feld is
E
=
k
C
q
r
2
.
The magnitudes of the electric Felds are
E
1
= 8
.
99
×
10
9
N
·
m
2
/
C
2
×
1
.
72
×
10
−
8
C
(0
.
386 m)
2
= 1037
.
8 N
/
C
,
and
E
2
= 8
.
99
×
10
9
N
·
m
2
/
C
2
×
6
.
29
×
10
−
8
C
(0
.
386 m)
2
= 3795
.
21 N
/
C
.
If the 17
.
2
×
10
−
9
C charge is the left one and
the positive direction is to the right,
E
net
=
E
1
−
E
2
= 1037
.
8 N
/
C
−
3795
.
21 N
/
C
=
−
2757
.
41 N
/
C
b
v
E
net
b
=
2757
.
41 N
/
C
.
The electric Feld at midpoint between the
charges +17
.
2
×
10
−
9
C and +62
.
9
×
10
−
9
C
is 2757
.
41 N
/
C toward the +17
.
2
×
10
−
9
C
charge.
002
10.0 points
A charge of +2
.
20
×
10
−
9
C is placed at the
origin, and another charge of +6
.
50
×
10
−
9
C
is placed at
x
= 1
.
7 m.
±ind the point (coordinate) between these
two charges where a charge of +3
.
80
×
10
−
9
C should be placed so that the net electric
force on it is zero. The Coulomb constant is
8
.
98755
×
10
9
N
·
m
2
/
C
2
.
Correct answer: 0
.
625258 m.
Explanation:
Let :
q
1
= 2
.
20
×
10
−
9
C at the origin
,
q
2
= 6
.
50
×
10
−
9
C
,
q
3
= 3
.
80
×
10
−
9
C
,
and
x
2
= 1
.
7 m
.
q
3
is between
q
1
and
q
2
, so if
r
1
,
3
=
P
, then
r
2
,
3
=
x
2
−
P
.
F
electric
=
k
C
q
1
q
2
r
2
The net electric force is zero, so
F
1
,
3
=
F
2
,
3
k
C
q
1
q
3
r
2
1
,
3
=
k
C
q
2
q
3
r
2
2
,
3
q
1
r
2
1
,
3
=
q
2
r
2
2
,
3
q
1
P
2
=
q
2
(
x
2
−
P
)
2
(
x
2
−
P
)
2
P
2
=
q
2
q
1
x
2
−
P
P
=
r
q
2
q
1
x
2
−
P
=
P
r
q
2
q
1
P
p
1 +
r
q
2
q
1
P
=
x
2
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View Full Document fullwood (tf5349) – Homework 1 Electrostatics – Catala – (12054)
2
P
=
x
2
1 +
r
q
2
q
1
=
1
.
7 m
1 +
r
6
.
5
×
10
−
9
C
2
.
2
×
10
−
9
C
=
0
.
625258 m
.
003
(part 1 of 2) 10.0 points
One gram of platinum has 3
.
09
×
10
21
atoms, and each platinum atom has 78 elec
trons.
How many electrons are contained in 6.00 g
of platinum?
Correct answer: 1
.
44612
×
10
24
electrons.
Explanation:
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This note was uploaded on 06/10/2010 for the course PHY 100 taught by Professor Pro.kiroshima during the Spring '10 term at Adelphi.
 Spring '10
 PRO.KIROSHIMA
 Electrostatics, Work

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