oldmidterm 02 – JYOTHINDRAN, VISHNU – Due: Oct 16 2007, 11:00 pm
Question 1, chap 26, sect 2.
part 1 of 1
10 points
A capacitor network is shown in the following figure.
4.39 µF
8.1 µF
a
b
12.1 µF
11 V
Explanation:
Let :
A coaxial cable with length ℓ has an inner
conductor that has a radius a and carries a
charge of Q. The surrounding conductor has
an inner radius b and a charge of −Q.
Assume the region between the conductors
Q
is air. The linear charge density λ ≡ .
ℓ
ℓ
radius = a
+Q
radius = b
−Q
What is the voltage across the 8.1 µF upper
right-hand capacitor?
Correct answer: 3.86629 V (tolerance ± 1
%).
C1
C2
C3
V
= 4.39 µF,
= 8.1 µF,
= 12.1 µF,
= 11 V.
C1 V1 = C2 V2 ,
and their voltages add up to V , voltage of the
battery
V1 + V2 = V
C2 V2
+ V2 = V
C1
C2 V2 + C1 V2 = V C1
V C1
V2 =
C1 + C2
(11 V)(4.39 µF)
=
4.39 µF + 8.1 µF
= 3.86629 V .
Question 2, chap 26, sect 1.
part 1 of 2
10 points
b
What is the electric field halfway between
the conductors?
1.
E=
2.
E=
3.
E=
4.
E=
5.
E=
6.
E=
7.
E=
8.
E=
9.
E=
10.
E=
and
Since C1 and C2 are in series they carry the
same charge
1
Q
π ǫ0 r
Q
π ǫ0 r 2
λ
2 π ǫ0 r 2
λ
correct
2 π ǫ0 r
Q
2 π ǫ0 r
Q
2 π ǫ0 r 2
λ
π ǫ0 r
λ
π ǫ0 r 2
λ
4 π ǫ0 r
Q
4 π ǫ0 r
Explanation:
The electric field of a cylindrical capacitor
is given by
E=
λ
.
2 π r ǫ0
oldmidterm 02 – JYOTHINDRAN, VISHNU – Due: Oct 16 2007, 11:00 pm
2
%).
Question 3, chap 26, sect 1.
part 2 of 2
10 points
Explanation:
What is the capacitance C of this coaxial
cable?
1.
C=
2.
C=
3.
C=
C=
4.
5.
C=
6.
C=
2ℓ
a
ke ln
b
ℓ
b
ke ln
a
ke ℓ
b
ln
a
ℓ
a
ke ln
b
2ℓ
b
ke ln
a
ℓ
correct
b
2 ke ln
a
C2
E
C4
d
C1
a
Question 4, chap 26, sect 3.
part 1 of 4
10 points
37.6 µF
Four capacitors are connected as shown in
the figure.
c
µF
2
17.
62.5 µF
b
A good rule of thumb is to eliminate junctions connected by zero capacitance.
.
b
2 ke ln
a
d
and
C3
ℓ
b
4
79.
= 17.2 µF ,
= 37.6 µF ,
= 62.5 µF ,
= 79.4 µF ,
= 97.3 V .
c
a
C2
C=
97.3 V
C1
C2
C3
C4
E
C1
Explanation:
a
Let :
µF
Find the capacitance between points a and
b of the entire capacitor network.
Correct answer: 120.077 µF (tolerance ± 1
C4
b
C3
Q
The definition of capacitance is C ≡ .
V
The series connection of C2 and C3 gives
the equivalent capacitance
1
C23 =
1
1
+
C2 C3
C2 C3
=
C2 + C3
(37.6 µF) (62.5 µF)
=
37.6 µF + 62.5 µF
= 23.4765 µF .
The total capacitance Cab between a and b can
be obtained by calculating the capacitance in
the parallel combination of the capacitors C1 ,
C4 , and C23 ; i.e.,
Cab = C1 + C4 + C23
= 17.2 µF + 79.4 µF + 23.4765 µF
= 120.077 µF .
oldmidterm 02 – JYOTHINDRAN, VISHNU – Due: Oct 16 2007, 11:00 pm
Question 5, chap 26, sect 3.
part 2 of 4
10 points
What is the charge on the 37.6 µF centeredupper capacitor?
Correct answer: 2284.27 µC (tolerance ± 1
%).
Explanation:
The voltages across C2 and C3 , respectively,
(the voltage between a and b) are Vab = V23 =
97.3 V , and we have
Q23 = Q3 = Q2
= Vab C23
= (97.3 V)(23.4765 µF)
= 2284.27 µC .
Question 6, chap 26, sect 3.
part 3 of 4
10 points
A dielectric with dielectric constant 2.61 is
inserted into the 62.5 µF capacitor (lowercentered capacitor) while the battery is connected.
What is the charge on the 79.4 µF lowerright capacitor?
Correct answer: 7725.62 µC (tolerance ± 1
%).
Explanation:
Since the battery is still connected, the voltage will remain the same. Thus the charge is
simply
Q4 = Vab C4
= (97.3 V) (79.4 µF)
= 7725.62 µC .
Question 7, chap 26, sect 3.
part 4 of 4
10 points
If the battery is removed before the dielectric in the above question is inserted, what
will be the charge on the 79.4 µF lower-right
capacitor?
Correct answer: 7295.45 µC (tolerance ± 1
%).
3
Explanation:
After the battery is removed, as the dielectric is inserted into C3 , there will be a
redistribution of charge.
Note: The total charge is unchanged, so
Qab = Q′ab .
The primed quantities correspond to those
after the insertion of the dielectric. Before the
battery was disconnected,
Qab = Vab Cab
C2 C3
= Vab C1 + C4 +
.
C2 + C3
from Part 1. After the battery was disconnected and the dielectric was inserted,
′
Vab
=
Qab
′ .
Cab
The equivalent capacitance of the two capacitors C2 and C3 is now
′
C23
=
C2 κ C3
,
C2 + κ C3
′
so the new total equivalent capacitance Cab
is
C2 κ C3
C2 + κ C3
= 17.2 µF + 79.4 µF
(37.6 µF) (2.61)(62.5 µF)
+
37.6 µF + (2.61)(62.5 µF)
= 127.157 µF .
′
Cab
= C1 + C4 +
′
The new voltage Vab
between a and b is therefore
Qab
′
Vab
= ′
Cab
Cab
= Vab ′ ,
Cab
since Q′ab = Qab = Vab Cab . The new charge
Q′4 on C4 is now
′
Q′4 = Vab
C4
Cab
= Vab ′ C4
Cab
(120.077 µF)
(79.4 µF)
= (97.3 V)
(127.157 µF)
= 7295.45 µC .
oldmidterm 02 – JYOTHINDRAN, VISHNU – Due: Oct 16 2007, 11:00 pm
part 1 of 1
Question 8, chap 26, sect 1.
part 1 of 1
10 points
A parallel-plate capacitor is charged by connecting it to a battery.
If the battery is disconnected and the separation between the plates is increased, what
will happen to the charge on the capacitor
and the electric potential across it?
1. The charge remains fixed and the electric
potential increases. correct
2. The charge decreases and the electric potential increases.
The resistance R of this component is
1. R =
2. R =
3. R =
5. R =
4. The charge and the electric potential remain fixed.
6. R =
5. The charge remains fixed and the electric
potential decreases.
7. R =
6. The charge increases and the electric potential decreases.
8. R =
9. R =
ρa
πℓ b2
2ρ ℓ
π b2
ρℓ
π a2
ρℓ
π b2
ρ ℓ b2
π a4
π b2 ρ
ℓ
ρ ℓ a2
π b4
ρℓ
correct
2
π (b − a2 )
2πρ ℓ
2
(b − a2 )
π b2 − a2 ρ
ℓ
7. The charge decreases and the electric potential remains fixed.
10. R =
8. The charge increases and the electric potential remains fixed.
Explanation:
By definition
9. The charge and the electric potential increase.
Explanation:
Charge is conserved, so it must remain constant since it is stuck on the plates. With the
battery disconnected, Q is fixed.
A
d
A larger d makes the fraction smaller, so C is
Q
smaller. Thus the new potential V ′ = ′ is
C
larger.
10 points
A resistor is made from a hollow cylinder
of length l, inner radius a, and outer radius b.
The region a < r < b is filled with material of
resistivity ρ. Current runs along the axis of
the cylinder.
4. R =
3. The charge and the electric potential decrease.
4
R=ρ
ℓ
A
ℓ
− π a2
ρℓ
.
=
2
π (b − a2 )
=ρ
π b2
C=ǫ
Question 9, chap 27, sect 2.
Question 10, chap 28, sect 1.
part 1 of 1
10 points
The emf of a battery is E = 11 V . When the
battery delivers a current of 0.7 A to a load,
the potential difference between the terminals
of the battery is 9 V volts.
oldmidterm 02 – JYOTHINDRAN, VISHNU – Due: Oct 16 2007, 11:00 pm
Find the internal resistance of the battery.
Correct answer: 2.85714 Ω (tolerance ± 1
%).
Explanation:
Given : E = 11 V ,
Vload = 9 V , and
I = 0.7 A .
Using Eq. (1), at room temperature; e.g.,
R1 = 14 Ω and ∆T1 = 25◦ C
R1
R0 =
(2)
[1 + α ∆T1 ]
14 Ω
=
1 + [0.004 (◦ C)−1 ] (25◦ C)
= 12.7273 Ω .
Using Eq. (1), with resistance R2 ,
The potential difference across the internal
resistance is E − Vload , so the internal resistance is given by
E − Vload
I
11 V − 9 V
=
0.7 A
= 2.85714 Ω .
r=
Question 11, chap 27, sect 3.
part 1 of 1
10 points
An incandescent bulb has a resistance of
14 Ω when it is at room temperature (25◦ C)
and 300 Ω when it is hot and delivering light
to the room. The temperature coefficient
of resistivity of the filament is 0.004 (◦ C)−1 ,
where the base resistance R0 is determined at
0◦ C.
What is the temperature of the bulb when
in use?
Correct answer: 5642.86 ◦ C (tolerance ± 1
%).
R2
−1
R0
∆T2 =
α
300 Ω
−1
12.7273
Ω
=
0.004 (◦ C)−1
= 5642.86◦ C .
(3)
Or directly substituting Eq. (1) into (2), we
have
R2
−1
R0
∆T2 =
α
R2
[1 + α ∆T1 ] − 1
R1
=
α
R2 + α R2 ∆T1 − R1
=
α
R2
R2 − R1
+
∆T1
(4)
=
α R1
R1
300 Ω − 14 Ω
=
[0.004 (◦ C)−1 ] (14 Ω)
300 Ω
(25◦ C)
+
14 Ω
= 5642.86◦C .
Question 12, chap 28, sect 4.
part 1 of 1
10 points
Explanation:
The equivalent resistance of the circuit in
the figure is Req = 55.0 Ω .
15 Ω
68 Ω
Let : R1 = 14 Ω ,
R2 = 300 Ω , and
α = 0.004 (◦ C)−1 .
68 Ω
15 Ω
Temperature dependence of resistance is
RT = R0 [1 + α (T − 0◦ C)]
= R0 [1 + α ∆T ] .
5
R
(1)
E
S
oldmidterm 02 – JYOTHINDRAN, VISHNU – Due: Oct 16 2007, 11:00 pm
Find the value of R.
Correct answer: 13.5 Ω (tolerance ± 1 %).
The circuit has been connected as shown in
the figure for a “long” time.
Explanation:
R2
E
Let :
R1
R2
R3
R4
Req
= 68.0 Ω ,
= 15.0 Ω ,
= 15.0 Ω ,
= 68.0 Ω ,
= 55.0 Ω .
Req,p
=
48 Ω
90 V
S
S
What is the magnitude of the electric potential across the capacitor?
Correct answer: 32 V (tolerance ± 1 %).
and
Basic Concepts: For resistors in parallel,
1
6Ω
18 µF
R1
R
48 Ω
42 Ω
R4
Explanation:
t
R2
R1
a
It
R3
1
1
+
Ra Rb
E
For resistors in series,
b
It
C
R3
6
R4
Ib
Ib
S1
b
Req,s = Ra + Rb
Let :
R12 = R1 + R2
= 68 Ω + 15 Ω
= 83 Ω , and
R34 = R3 + R4
= 15 Ω + 68 Ω
= 83 Ω , and
−1
1
1
R1234 =
+
R12 R34
−1
1
1
+
=
83 Ω 83 Ω
= 41.5 Ω , and
R = Req − R1234
= 55 Ω − 41.5 Ω
R1
R2
R3
R4
C
= 42 Ω ,
= 48 Ω ,
= 6 Ω,
= 48 Ω , and
= 18 µF .
After a “long time” implies that the capacitor C is fully charged and therefore the
capacitor acts as an open circuit with no current flowing to it. The equivalent circuit is
It
It
a
R1
R3
Ib
R2
b
R4
Ib
= 13.5 Ω .
Rt = R1 + R2 = 42 Ω + 48 Ω = 90 Ω
Question 13, chap 28, sect 7.
part 1 of 2
10 points
Rb = R3 + R4 = 6 Ω + 48 Ω = 54 Ω
oldmidterm 02 – JYOTHINDRAN, VISHNU – Due: Oct 16 2007, 11:00 pm
It =
Ib =
E
90 V
=
=1A
Rt
90 Ω
7
where
Rℓ = R1 + R3 = 42 Ω + 6 Ω = 48 Ω,
Rr = R2 + R4 = 48 Ω + 48 Ω = 96 Ω
90 V
E
= 1.66667 A
=
Rb
54 Ω
and
Across R1
E 1 = It R 1
= (1 A) (42 Ω)
= 42 V .
Req =
E 3 = Ib R 3
= (1.66667 A) (6 Ω)
= 10 V .
|EC | = 32 V .
Question 14, chap 28, sect 7.
part 2 of 2
10 points
If the battery is disconnected, how long
does it take for the capacitor to discharge to
1
Et
= of its initial voltage?
E0
e
Correct answer: 576 µs (tolerance ± 1 %).
Explanation:
With the battery removed, the circuit is
Ir
R2
C
r
The equation for discharge of the capacitor is
Qt
= e−t/τ , or
Q0
1
Et
= e−t/τ = .
E0
e
Taking the logarithm of both sides, we have
t
1
− = ln
τ
e
t = −τ (− ln e)
= −(576 µs) (−1)
= 576 µs .
Question 15, chap 28, sect 4.
part 1 of 1
10 points
A network below consists of with three batteries, each having an internal resistance, and
five resistors.
13 V
9Ω
5Ω
1Ω
R4
Iℓ
Ir
23 V
Ieq
Req
a
C
−1
τ ≡ Req C = (32 Ω) (18 µF) = 576 µs .
EC = E3 − E1
= 10 V − 42 V
= −32 V
R1
R3
−1
Therefore the time constant τ is
Since E1 and E3 are “measured” from the same
point “a”, the potential across C must be
ℓ
1
1
+
Rℓ Rr
1
1
+
=
48 Ω 96 Ω
= 32 Ω .
Across R3
Iℓ
b
6Ω
32 V
1Ω
5Ω
1Ω
4Ω
oldmidterm 02 – JYOTHINDRAN, VISHNU – Due: Oct 16 2007, 11:00 pm
Find the magnitude of the potential difference between points a and b.
Correct answer: 0.961538 V (tolerance ± 1
%).
Explanation:
E1
R1
r1
i
a
i
b
R2
R3
E2
r2
R5
E3
r3
R4
i
i
Let : E1
E2
E3
R1
R2
R3
R4
R5
r1
r2
r3
= 13 V ,
= 23 V ,
= 32 V ,
= 9 Ω,
= 6 Ω,
= 5 Ω,
= 4 Ω,
= 5 Ω,
= 1 Ω,
= 1 Ω , and
= 1 Ω.
Basic Concepts: Kirchhoff’s Laws:
X
V = 0 around a closed loop.
X
8
We have
E1 − E3
r1 + R 1 + R 2 + r3 + R 4 + R 3
E1 − E3
=
R1234
13 V − 32 V
=
26 Ω
= −0.730769 A .
i=
We can then find the magnitude of the potential difference between a and b, |Vab |, by using
the top loop; i.e.,
Vab = E1 − E2 − i (R1 + r1 + R3 )
= (13 V) − (23 V)
− (−0.730769 A) (9 Ω + 1 Ω + 5 Ω)
= −0.961538 V
|Vab | = 0.961538 V ,
where the current through r2 and R5 is zero.
Question 16, chap 28, sect 7.
part 1 of 1
10 points
In the circuit shown, the capacitor is initially uncharged. At t1 = 0, the switch S is
moved to position “a”.
R2
C
R1
S b
V0
a
I = 0 at a circuit junction.
Solution: There is no current flow between
a and b. Therefore, applying the loop rule to
the outside path in the figure above, we have
E1 − E3 − i (r1 + R1 + R2 + r3 + R4 + R3 ) = 0 ,
Since
R1234 = 1 Ω + 9 Ω + 6 Ω
+1 Ω+4Ω+5Ω
= 26 Ω ,
Find VR1 , the voltage drop across R1 , as a
function of time t1 .
1. VR1 = V0 e−t1 /(R2 C)
o
n
−t1 /(R1 C)
2. VR1 = V0 1 − e
o
n
t1 /[(R1 +R2 ) C]
3. VR1 = V0 1 − e
4. VR1 = V0 e−[t1 (R1 +R2 )]/R1 R2 C
5. VR1 = V0 et1 /[(R1 +R2 )C]
oldmidterm 02 – JYOTHINDRAN, VISHNU – Due: Oct 16 2007, 11:00 pm
7. VR1 = V0
n
[−t1 (R1 +R2 )]/R1 R2 C
1−e
−t1 /(R2 C)
1−e
o
o
b = +ı̂
6. F
b = −k̂
7. F
Explanation:
Basic Concepts:
Charged Particle:
8. VR1 = V0 e−t1 /(R1 C) correct
Explanation:
For an “RC” circuit,
~ = q ~v × B
~
F
I = I0 e−t1 /(RC)
V0 −t1 /(R1 C)
=
e
.
R1
Right-hand rule for cross-products.
~
b ≡ F ; i.e., a unit vector in the F direcF
~k
kF
tion.
~ = q ~v × B.
~
Solution: The force is F
~ = B −k̂ ,
B
~v = v +k̂ , and
Since I R1 = VR1 ,
VR1 = V0 e−t1 /(R1 C) .
Question 17, chap 29, sect 1.
part 1 of 1
10 points
A positively charged particle moving parallel to the z-axis enters a magnetic field (pointing toward the left-hand side of the page), as
shown in the figure below.
x
~
B
y
+q
z
v
Figure: ı̂ is in the x-direction, ̂ is
in the y-direction, and k̂ is in the
z-direction.
What is the initial direction of deflection?
~ = 0 ; no deflection correct
2. F
b = +k̂
3. F
b = −̂
4. F
b = −ı̂
5. F
q > 0 , therefore,
~
~
F = +|q| ~v × B
h i
= +|q| v B +k̂ × −k̂
= +|q| v B (0)
b = 0 no deflection .
F
This is the second of eight versions of the
problem.
Question 18, chap 29, sect 1.
part 1 of 1
10 points
~
B
b = +̂
1. F
Magnetic Force on a
A wire carrying a current 20 A has a length
0.1 m between the pole faces of a magnet at
an angle 60 ◦ (see the figure). The magnetic
field is approximately uniform at 0.5 T. We
ignore the field beyond the pole pieces.
I
B
θ
ℓ
6. VR1 = V0
n
9
What is the force on the wire?
Correct answer: 0.866025 N (tolerance ± 1
%).
oldmidterm 02 – JYOTHINDRAN, VISHNU – Due: Oct 16 2007, 11:00 pm
Explanation:
10
y
x
d
we use F = I ℓ B sin θ, so
F = I ℓ B sin θ
= (20 A) (0.1 m) (0.5 T) sin 60 ◦
= 0.866025 N .
ℓ
Let : I = 20 A ,
ℓ = 0.1 m ,
θ = 60 ◦ , and
B = 0.5 T .
→ I →
← axis of rotation
~
B
~
B
Let : d = 0.39 m ,
ℓ = 0.39 m ,
B = 0.28 T ,
I = 4 A.
and
Only the right side of the loop contributes to
the torque. By definition, the torque is
Question 19, chap 30, sect 3.
part 1 of 1
10 points
A square loop of wire carries a current and
is located in a uniform magnetic field.
The left side of the loop is aligned and
attached to a fixed axis (dashed line in figure).
y
x
0.39 m
→ 4A →
← axis of rotation
~ = 0.28 T
B
0.39 m
~ = 0.28 T
B
When the plane of the loop is parallel to the
magnetic field in the position shown, what is
the magnitude of the torque exerted on the
loop about the axis of rotation, which is along
the left side of the square as indicated by the
dashed line in the figure?
Correct answer: 0.170352 N m (tolerance ± 1
%).
Explanation:
~k
τ = k~d × F
= d×I ℓB
= (0.39 m) (4 A) (0.39 m) (0.28 T)
= 0.170352 N m .
Question 20, chap 29, sect 2.
part 1 of 1
10 points
What is the radius of the smallest possible
circular orbit that a 0.95 MeV proton can
have in a 2 T magnetic field?
The mass of a proton is 1.67 × 10−27 kg and
its charge is 1.609 × 10−19 C.
Correct answer: 0.0702145 m (tolerance ± 1
%).
Explanation:
Let : m = 1.67 × 10−27 kg ,
Q = 1.609 × 10−19 C ,
B = 2 T , and
E = 0.95 MeV .
1
The energy of a proton is E = m v 2 , so
2
r
2E
v=
m
oldmidterm 02 – JYOTHINDRAN, VISHNU – Due: Oct 16 2007, 11:00 pm
=
s
2 (0.95 MeV) (1.609 × 10−19 J/eV)
(1 × 10−6 MeV/eV) (1.67 × 10−27 kg)
= 1.353 × 107 m/s .
If the proton is shot into the magnetic field
with a velocity at right angles to the direction
of the field, we will get the smallest radius
mv
Bq
(1.67 × 10−27 kg) (1.353 × 107 m/s)
=
(2 T) (1.609 × 10−19 C)
r=
= 0.0702145 m .
Consider the setup of a velocity selector for
~ is pointing
the case where the electric field E
downward along y axis.
y
v
b = √1 (̂ + k̂)
8. B
2
1
b = √ (−̂ − k̂)
9. B
2
b
10. B = +̂
Explanation:
The negatively charged particle passes
through the selector undeflected when the
electric force is equal and opposite to the magnetic force. Since the electric field is pointing
in the negative y direction, and the particle’s
charge is negative, the particle feels an electric
force in the positive y direction.
~ E = |q| E ̂
F
Question 21, chap 29, sect 2.
part 1 of 1
10 points
−
11
where E is the magnitude of the electric field.
We must choose the magnetic field such that
the magnetic force is in the negative y direction. The general equation for the magnetic
force is given by
~ B = q (~v × B)
~
F
E
O
x
z
Choose the direction of the magnetic field
~ such that a negatively charged particle,
B
moving at an appropriate speed in the positive
~ and B
~
x-direction (ı̂), passes through the E
region undeflected.
b = −ı̂
1. B
b = √1 (̂ − k̂)
2. B
2
b
3. B = −̂
b = +k̂
4. B
b = +ı̂
5. B
b = −k̂ correct
6. B
b = √1 (−̂ + k̂)
7. B
2
~ B to point opposite to F
~ E , i.e., in
We want F
the (−̂) direction to have a chance of cancelling the net force. Considering only the
unit vectors along which the above vectors
point, we must satisfy
−̂ = −(ı̂ × r̂)
(1)
where r̂ denotes the unknown direction in
~ must point. Since
which B
ı̂ × (−k̂) = ̂
we see that equation (1) for r̂ is satisfied by
r̂ = −k̂
so the magnetic field must point into the page.