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221
HOMEWORK
3:
SOLUTIONS
Q22.14 ANSWER: a) Yes, just use the Gauss’s theorem that includes TOTAL charge; b) No – the same reason as a)
22.14.
IDENTIFY:
Apply the results of Examples 22.9 and 22.10.
SET UP:
2
q
Ek
r
outside the sphere. A proton has charge +
e
.
EXECUTE:
(a)
19
9
2
2
21
2
15
2
92(1.60 10
C)
(8.99 10 N m /C )
2.4 10 N/C
(7.4 10
m)
q
r
(b)
For
10
1.0 10
m
r
,
2
15
21
13
10
7.4 10
m
(2.4 10 N/C)
1.3 10 N/C
1.0 10
m
E
(c)
0
E
, inside a spherical shell.
EVALUATE:
The electric field in an atom is very large.
22.41.
IDENTIFY:
First make a freebody diagram of the sphere. The electric force acts to the left on it since the electric
field due to the sheet is horizontal. Since it hangs at rest, the sphere is in equilibrium so the forces on it add to zero,
by Newton’s first law. Balance horizontal and vertical force components separately.
SET UP:
Call
T
the tension in the thread and
E
the electric field. Balancing horizontal forces gives
T
sin
=
qE.
Balancing vertical forces we get
T
cos
=
mg
. Combining these equations gives tan
=
qE/mg
, which means that
= arctan(
qE/mg
). The electric field for a sheet of charge is
0
2.
E
P
EXECUTE:
Substituting the numbers gives us
72
4
12
2
2
0
2.50 10 C/m
1.41 10 N/C
2
2 8.85 10
C /N m
E
. Then
84
22
5.00 10 C 1.41 10 N/C
arctan
19.8
2.00 10 kg 9.80 m/s
EVALUATE:
Increasing the field, or decreasing the mass of the sphere, would cause the sphere to hang at a larger
angle.
Q23.10: ANSWER: NO,  only potential difference is physical quantiry. We can always add an arbitrary constant to a
potential.
23.18.
IDENTIFY:
Apply
.
a
a
b
b
K
U
K
U
SET UP:
Let
1
3.00 nC
q
and
2
2.00 nC.
q
At point
a
,
12
0.250 m
aa
rr
. At point
b
,
1
0.100 m
b
r
and
2
0.400 m
b
r
. The electron has
qe
and
31
e
9.11 10
kg
m
.
0
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This note was uploaded on 02/11/2012 for the course PHYSICS 222 taught by Professor Ogilvie during the Fall '05 term at Iowa State.
 Fall '05
 Ogilvie
 Charge, Work

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