PHY2049 Fall 2009
Profs. A. Petkova, A. Rinzler, S. Hershfield
Final Exam Solution
1. A charge +3
.
00
q
lies fixed at the origin and a second charge of

2
.
00
q
lies fixed in the xy plane at
x
=

3
.
00,
y
=

4
.
00.
The x component of the force experienced by the

2
.
00
q
charge due to the charge at the origin is?
Answer:
0
.
144
kq
2
Solution:
The distance between the charges is 5 =
√
3
2
+ 4
2
. The magnitude of the force between the two charges is
k
(2
q
)(3
q
)
/
5
2
. Since the

2
q
charge is attracted to the +3
q
charge, the xcomponent of the force on the

2
q
charge is
positive. To get the xcomponent we need to multiply by the cosine of the angle between the force vector and the xaxis.
The cosine is the adjacent over the hypotenuse and equal to 3
/
5. Thus, the xcomponent of the force is (3
/
5)
k
(2
q
)(3
q
)
/
5
2
.
2. A circular insulating ring of radius
r
lies in the xy plane centered on the origin. The parts of the ring in the negative y
half plane are uncharged. The parts of the ring in the positive y half plane are uniformly charged with a total charge
Q
.
The electric field at the origin is:
Answer:

2
kQ
πr
2
ˆ
j
Solution:
First, by symmetry the net electric field is in the negative ydirection if we take
Q
to be positive. The charge
per unit length of the half ring is
λ
=
Q/
(
πr
). Take a small element of charge,
dq
=
λrdθ
= (
Qπ
)
dθ
. This charge produces
a field of magnitude
kdq/r
2
at the origin. Letting the angle the field makes with the yaxis be
θ
, the ycomponent of the
electric field at the origin is
E
y
=

integraldisplay
π/
2
−
π/
2
k
r
2
Q
π
cos(
θ
)
dθ
=

kQ
πr
2
(sin(
π/
2)

sin(

π/
2))
.
If
Q
is negative, then this formula still works, and the evaluated
E
y
is in positive ydirection because then

Q >
0.
3. In the figure the capacitances of the series capacitors are equal.
The
voltage drop across
C
4
is 12 V and the charge on it is 6.0 nC. The
charge on the entire network in 9.0 nC.
The capacitance of
C
2
is?
Answer:
0.75 nF
Solution:
The voltage across
C
1
,
C
2
, and
C
3
in series is 12 V. Because these capacitors are in series and equal, their
effective capacitance is
C
2
/
3. Since the net charge on the network is 9 nC and the charge on
C
4
is 6 nC, the charge on
the series resistors is 3 nC. Consequently,
C
2
/
3 = 3
nC/
12
V
, which implies that
C
2
= 9
/
12 nF.
4. The resistance measured between the ends of a metal wire having a circular crosssection is R. The wire diameter is
halved by going through a series of rollers that preserves the volume of the wire (i.e. the wire gets correspondingly longer
as its diameter shrinks with no loss of material). The resistance between the ends of the wire is now:
Answer:
16R
Solution:
The volume of the wire is
πr
2
L
, where
r
is the radius of the wire and
L
is the length of the wire. If the radius
is decreased by a factor of 2, then the length must be increased by a factor of 4 in order to keep the volume constant.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Any
 Charge, Force, Electric charge

Click to edit the document details