ID: A

ID: A
UCSD Physics 2B
Answer Section
Summer Session
SAMPLE Final Exam
MULTIPLE CHOICE
1. ANS: C
F=k
q1q2
r2
ÊÁ
ˆÊ
ˆ
Á 45.0 × 10−9 C ˜˜ ÁÁ 63.0 × 10 −9 C ˜˜
Ë
Ë
¯
¯
Ê
ˆ
= ÁÁ 9.0 × 10 9 Nm2 / C 2 ˜˜
= 4.08 × 10 −2 N
2
Ë
¯
ÊÁ
−3 ˆ
Á 25.0 × 10 m˜˜
Ë
¯
PTS: 1
2. ANS: E
+
1e
ˆ
Ê
# e = ÁÁ 410 × 10 −9 C ˜˜ Ê
= 2.56 × 10 +12
Ë
¯ ÁÁ 1.60 × 10 −19 C ˆ˜˜
¯
Ë
PTS: 1
3. ANS: B
By definition E =
F
1200N
N
=
= 5.2 × 10 4
q 23 × 10−3 C
C
PTS: 1
4. ANS: A
The absolute potential at distance r from a point charge q is
q
V = k which varies inversely (not inverse square) with distance
r
PTS: 1
5. ANS: A
Capacitors in parallel add C PAR = C 1 + C 2 + ë
PTS: 1
6. ANS: B
Resistors in parallel obey the formula
R PAR =
1
1
1
=
+
+ ë and, when all N resistors have the same value,
R PAR R 1 R 2
1
R
N
PTS: 1
1
ID: A
7. ANS: C
Current is equal to charge per unit time, but we must convert electron charge into Coulombs
I=
Q 475 × 1016 e −
=
t
1 min
ˆ
ÊÁ
ÁÁ 1.6 × 10 −19 C ˜˜˜ ÁÊÁ 1 min ˜ˆ˜
−2
˜˜ ÁÁ
˜
ÁÁ
˜˜ Á 60sec ˜˜ = 1.27 × 10 A = 12.7 mA
ÁÁ
1 e−
¯
¯Ë
Ë
PTS: 1
8. ANS: C
By definition
J=
I
I
12A
A
=
=
= 2.9 × 10 2 2
2
Area π R 2
ˆ
ÊÁ
m
π Á 3.6 × 10 −2 m˜˜
¯
Ë
PTS: 1
9. ANS: B
V
210V
R=
=
= 20Ω
I
10.5A
PTS: 1
10. ANS: A
ÊÁ
C ˆ˜
J ˆ˜ ÊÁ
E = P × t = V × I × t = ÁÁÁ 12.0 ˜˜˜˜ ÁÁÁÁ 52.0 ˜˜˜˜ (90 sec) = 56.2 kJ
ÁË
s¯
C ¯Ë
PTS: 1
11. ANS: D
The field at the center of a current loop is B =
μ0 I
2R
⇒I=
BD
μ0
PTS: 1
12. ANS: D
dΦ
NBA
emf =
=
so emf is inversely proportional to period.
dt
T
PTS: 1
13. ANS: B
V peak = 2V rms =
PTS:
14. ANS:
VP
=
VS
2 (440V ) = 622V
1
B
ÊÁ 18,000 ˆ˜
Ns
NP
˜˜ = 150,000V
⇒ VS = VP
= (2500V ) ÁÁÁÁ
˜
ÁË 300 ˜˜¯
NS
Np
PTS: 1
2
ÊÁ
ˆÊ
ˆ
Á 3.28 × 10 −4 T ˜˜ ÁÁ 17.0 × 10 −3 m˜˜
Ë
¯Ë
¯
=
= 4.43A
ÊÁ
ˆ˜
−6
Á 1.26 × 10 H / m˜
Ë
¯
ID: A
15. ANS: D
1
ω=
=
LC
PTS: 1
16. ANS: A
1
= 3.22 × 10 3 rad/sec
ÊÁ
ˆ˜ ÊÁ
−3
−3 ˆ
Á 0.370 × 10 H ˜ Á 0.260 × 10 F ˜˜
Ë
¯Ë
¯
Ê
Ë
ˆ
¯
τ = RC = (160Ω) ÁÁ 55 × 10−3 F ˜˜ = 8.8 s
PTS: 1
17. ANS: E
emf = −L
dI
= − (6.5H ) (−50A / s) = 325V = 0.33 kV
dt
PTS: 1
18. ANS: E
Definition
PTS: 1
19. ANS: A
Definition
PTS: 1
NUMERIC RESPONSE
1. ANS: U =
1
1
CV 2 =
2
2
ÊÁ
ˆÊ
ˆ2
Á 3.0 × 10 −6 C ˜˜ ÁÁ 5.0 × 10 3 V ˜˜ = 37.5J
Ë
¯Ë
¯
PTS: 1
2. ANS:
This circuit can be reduced to two series capacitors. The two in parallel add, to give C 23 = 2μF . Hence, since
the charge on C1 must be the same as the charge on C23 (series)
C1
V 23
1
=
= . Also, we require that V = V 1 + V 23 , so that
we must have C 1 V 1 = C 23 V 23 . In other words,
V1
C 23 2
V 23 = 4V .
PTS: 1
Ê
ˆÊ
ˆ
3. ANS: τ = p × E = E sin ϑ = ÁÁ 25 × 10 −9 C ⋅ m˜˜ ÁÁ 3.0 × 10 −6 N / C ˜˜ sin (65) = 6.8 × 10 −14 N ⋅ m
Ë
¯Ë
¯
PTS: 1
Ê
ˆÊ
ˆ
4. ANS: V = Ed = ÁÁ 23 × 10 3 N / C ˜˜ ÁÁ 2.4 × 10 −2 m˜˜ = 5.5 × 102 V
¯Ë
¯
Ë
PTS: 1
3
ID: A
ÁÊ 1 1 ˜ˆ
5. ANS: V = kQ ÁÁÁÁ − ˜˜˜˜
Ëa b¯
PTS: 1
6. ANS:
1
1
1
1
1
=
+
=
+
= 9.8 × 10 −2 Ω −1
R par R 1 R 2 25.0Ω 17.2Ω
P=
V2
2Ê
ˆ
= (50.0V ) ÁÁ 9.8 × 10−2 Ω −1 ˜˜ = 245W
Ë
¯
R
PTS: 1
7. ANS: The circuit can be seen a two pairs of 2-series resistors in parallel. The points opposite A & B
are connected across the battery, so both top & bottom pairs have 100V across them. The series
resistors are voltage dividers. The point B is
20Ω
× 100 V = 40 V above the left node. The point
20Ω + 30Ω
A is similarly 60 V above the left node. The difference is 20 V.
PTS: 1
8. ANS: Φ = π R 2 E
PTS: 1
9. ANS: E =
kq
2a
2
1
2
PTS: 1
10. ANS: V 2 = 0.25V 1
PTS: 1
11. ANS:
a ) I1 =
V
since capacitor C is a dead-short (perfect wire) when Q = 0. In other words, resistor R2 is also
R1
shorted and therefore not part of the circuit.
b ) I1 = I2 =
V
since capacitor C is open when fully charged. In other words, capacitor C is not part of
R1 + R2
the circuit anymore.
PTS: 1
4
ID: A
12. ANS:
Draw an Amperean Loop of width L around the slab. The right hand indicates the direction of the
field above and below (none in the vertical direction - recall the solenoid derivation).
ÿ B • dl = μ I
0 enc
⇒B=
È
˘
⇒ 2BL = μ0 (J × Area ) = μ0 ÍÍÍÎ J (Ld ) ˙˙˙˚
1
μ Jd
2 0
PTS: 1
13. ANS:
ˆ
Ê
ÁÁ 1.26 × 10 −6 H / m˜˜ (75A)
μ0 I r 2
¯
Ë
B=
=
Ê
ˆ
2π r R 2
Á
(6.28) Á 3.5 × 10−2 m˜˜
Ë
¯
2
ÁÊÁ 3.5 × 10−2 m˜ˆ˜
¯
Ë
= 2.11 × 10 −4 T = 0.21mT
2
ÊÁ
ˆ
Á 5.0 × 10−2 m˜˜
Ë
¯
PTS: 1
14. ANS: U =
1 2 1
LI =
2
2
ÊÁ
2
ˆ
Á 35.0 × 10 −3 C ˜˜ (52.0A) = 47.3J
Ë
¯
PTS: 1
15. ANS:
Note: the units given in V ÊÁË x,y,z ˆ˜¯ must be Volts. Hence the "y" term, for example, must include units which
cancel out the units generated by y 3. Here we go . . .
E ÊÁË x,y,z ˆ˜¯ = −∇V =
ÊÁ ∂V
∂V
∂V ˆ˜˜
ÁÁ −
ÁÁ ∂x , − ∂y , − ∂z ˜˜˜
¯
Ë
ÊÁ
V
V
V ˆ˜
= ÁÁÁÁ −30x 2 , + 30y 2 3 , − 88z 3 4 ˜˜˜˜
m
m
m ¯
Ë
= −30x
V 8
V
V
i + 30y 2 3 8j − 88z 3 4 k8
2
m
m
m
(-30V/m)x i +(30V/m)y2 j +(-88V/m)z3 k
PTS: 1
5
ID: A
PROBLEM
1. ANS:
a ) to the right
total charge +Q
=
b) λ=
arc length
πa
c ) Let arc-length = s. Then dq = λ (ds) = λ (ad θ ) =
d ) dE = k
dq
a
2
r8 = k
+Q
π a2
e ) dE x = dE sin θ = k
∫
f ) E x = dE x =
=k
∫
π
0
k
ad θ =
+Q
π
dθ
d θ r8
+Q
π a2
+Q
πa
+Q
πa
2
sin θ d θ r8
sin θ d θ = k
+Q ÈÍ
˘
ÍÎ (−cos π ) − (−cos 0) ˙˙˙˚
2 Í
πa
+Q ÈÍ
+Q
˘
ÍÎ (− −1) − (−1) ˙˙˙˚ = 2k
2 Í
πa
π a2
PTS: 1
2. ANS:
a) Let i = right, j = up and k = out. Then
Ê
ˆÊ
ˆ
F = qvB ÊÁÁ -j8 ˆ˜˜ = ÁÁ −1.60 × 10 −19 C ˜˜ ÁÁ 133 × 10 −2 m / s ˜˜ (1.20T ) ÊÁÁ -j8 ˆ˜˜ = 2.55 × 10 −19 N ÊÁÁ j8 ˆ˜˜
Ë ¯ Ë
Ë ¯
Ë ¯
¯Ë
¯
ˆ
−19 Ê
−2.55 × 10 N ÁÁ j8 ˜˜
−F
N
Ë ¯
=
= 1.596
b) −F = qE ⇒ E =
−19
q
C
1.60 × 10 C
N
Ê
ˆ
check: qE = −qvB ⇒ E = −vB = ÁÁ 133 × 10 −2 m / s ˜˜ (1.2T ) = 1.596
Ë
¯
C
ÊÁ
N ˆ˜ Ê
J
ˆ
c) V = Eh = ÁÁÁÁ 1.596 ˜˜˜˜ ÁÁ 33 × 10 −2 m˜˜ = 0.527
C
C ¯Ë
¯
Ë
PTS: 1
6
ID: A
3. ANS:
a ) Since the shell is a perfect conductor, the electric field must be zero inside. The only way that can happen
(draw a Gaussian surface inside the shell) is if there is a canceling charge +Q on the inner surface. Since the
shell itself is said to be neutral, there must be an opposite charge -Q on the outer surface.
b ) A Guassian surface drawn inside the inner surface encloses only the point charge -Q. Hence the field must
be E = −kQ / r 2
c ) A Guassian surface drawn within the shell itself contain the center point charge -Q
plus the inner
surface charge +Q and the two cancel. But we already know this since E = 0 inside a conductor.
d ) A Gaussian surface drawn outside this whole business contains net charge -Q since the shell itself is
neutral. Hence, the electric field has the same form as part (a).
e ) Since the electric field outside the shell is simply that of the point charge at the center, the potential must
be that as well, namely, V = −kQ / b
f ) This is a bit tricky - we know the potential at r = b. But to move a test charge through the conductor to the
inner surface r = a we do no work since the field inside is zero. Hence V (a ) = V (b) = −kQ / b
a
a
−kQ
E • dr =
dr OR we can simply note that
g ) This is even trickier - we must do the integral ΔV =
2
a/2
a/2 r
ÊÁ 1
ÊÁ 2 1 ˆ˜ −kQ
1 ˆ˜
− ˜˜˜˜ = −kQ ÁÁÁÁ − ˜˜˜˜ =
the potential difference is ΔV a,a / 2 = −kQ ÁÁÁÁ
a
Ëa/2 a¯
Ëa a¯
∫
∫
To get the absolute potential, we must add the potential at r = a to this:
ÁÊ 1 1 ˜ˆ
V (a / 2) = V (a) + ΔV a,a / 2 = −kQ / b − kQ / b = −kQ ÁÁÁÁ + ˜˜˜˜
Ëa b¯
PTS: 1
4. ANS:
a ) Use the right-hand-rule and you’ll see the net force is out of the page
b ) The angle between dL and B is θ
c ) Use dF = IdL × B = I (dL)B sin θ direction is always out of the page
d ) Integrate from θ = 0 to θ = 180°
PTS: 1
7
ID: A
5. ANS:
The hint tells us that we can treat the pair of slabs as if they were infinite sheets.
a ) The right-hand-rule says the field between the slabs is to the right
b ) Draw an Amperean Loop around either slab. Then
μ0 I
B • dL = μ0 I enclosed The field outside = 0, so Bw = μo I or B =
w
c ) The area is the height times the length (B points out the side)
Area = d × l
μ0 I
× (l d )
so the flux is Φ B = B • A =
w
d ) The inductance is just the flux divided by the current that made it - a measure of the device!
Φ μo ld
L=
=
I
w
∫
PTS: 1
8
ID: A
6. ANS:
Here the ‘height’is the length. The cross section looks like this:
b
a
a ) In order to find the resistance, note that the current must flow through s series of cyclindrical shells to get
from the inside out. Such a shell cross section would look like this:
h
R
A= 2πrh
2πr
dr
r
Laid out, it would be a rectangle of length h and width equal to the circumference 2π r and thickness d = dr
b ) The resistance of a rectangular slab is R = ρ
d
so that the differential resistance is just the that of the
A
dr
2π rh
c ) The total resistance is the sum of the resistances of all these slabs (which are in series) and this becomes an
b
ρ b dr
ρ
ÁÊ b ˜ˆ
dr
=
=
ln ÁÁÁÁ ˜˜˜˜
integral : R = dR = ρ
2π rh 2π h a r
2π h Ë a ¯
a
infinitesimal ‘slab’ in part ( a ) dR = ρ
∫
∫
∫
NOTE: if this problem is on the Final, it will un doubtedly have a function ρ (r) instead of a constant but,
even so, you’ll only have to integrate a function of one variable.
PTS: 1
9