Name:
Solution Key
ID No.:
Instructor:
Louisiana State University Physics 2102, Exam 2,
October 14, 2004
• Please, be sure to write your name, student ID number, and class instructor above.
• For the problems: Show your reasoning and your work.
• The questions are multiple choice. No partial credit on the questions.
• You may use scientific or graphing calculators, but you must derive and explain your answer fully
on paper so we can grade your work.
• Feel free to detach, use, and keep the formula sheet pages. No other reference material is allowed
during the exam.
• Good luck!
Question 1 [8 points]
Figure shows an open switch, a battery of potential difference V , and three identical uncharged capacitors of capacitance C.
1
2
3
series connection: → q1 = q2 = q3 = q
→ V = V1 + V2 + V3
V
+
−
C
C
C
condition of the problem: C1 = C2 = C3 = C
Therefore: V = 3qC, V1 = qC = V2 = V3 = V /3
r©© r
(a) (4 pts) When the switch is closed and the circuit reaches equilibrium what is the potential difference
across each capacitor? (Circle the right answer).
(i)
V1 = V
V2 = V /2
V3 = V /3
(ii)
V1 = V
V2 = 0
V3 = 0
[(iii)]
V1 = V /3
V2 = V /3
V3 = V /3
(b) (4 pts) When the switch is closed and the circuit reaches equilibrium what is the charge on the
left-hand plate of each capacitor? (Circle the right answer).
(i)
q1 > q2 > q3
[(ii)]
q1 = q2 = q3
(iii)
q1 = q3 > q2 = 0
2
Problem 1 [17 points]
In figure below assume that C1 = 10.0 µF, C2 = 20.0 µF, C3 = 20.0 µF, and E = 100 V.
q2
E
V1
V2
C2
Show all the steps to your final answers!
C1
C3
q1
q3
V3
The capacitors C2 and C3 are connected in series,
C2 C3
therefore: C23 =
C2 + C3
The capacitor C1 and the combination C23 are
connected in parallel, therefore, Ceq = C1 + C23 .
(a) (6 pts) Calculate the equivalent capacitance, Ceq , of the combination.
Ceq = C1 + C23 = C1 +
C2 C3
(20 µF)(20 µF)
= (10 µF) +
= 20 µF.
C2 + C3
40 µF
(b) (5 pts) Calculate the charge q1 .
q1 = C1 V1 = C1 E = (10 µF)(100 V) = 10−3 C.
(c) (6 pts) Calculate the potential energy U3 stored in the charged capacitor C3 .
One way to solve this:
The charge on the whole combination is: q = Ceq E = (20 µF)(100 V) = 2 ×10−3 C
The charges on both seriesly connected capacitors are equal, q2 = q3 , and each of them is equal to q − q1
because of the parallel connection. Therefore,
q3 = q − q1 = (2 − 1) × 10−3 = 1 × 10−3 C
The energy stored in the charged capacitor C3 is:
U3 =
q32
(1 × 10−3 C)2
=
= 0.025 J.
2C3
2(20 × 10−6 F)
Another way:
In this combination the battery, the C1 , and C23 are connected in parallel, therefore: E = V1 = V23
The C2 and C3 are in series, therefore: q2 = q3 = q23 → C2 V2 = C3 V3 = C23 V23 = C23 E
Choosing second part of the upper equality we have:
U3 =
C3 V32
(20 × 10−6 F)(50 V)2
=
= 0.025 J.
2
2
3
V3 =
(10 × 10−6 F)(100 V)
C23 E
=
= 50 V
C3
20 × 10−6 F
Question 2 [8 points]
The figure shows three cylindrical copper conductors along with their face areas and lengths.
L
A ¶³
2 µ´
A
(1)
1.5 L
³
´
L/2
A ¶³ ³
2 µ´ ´
(2)
(3)
(a) (4 pts) Rank the cylinders according to their resistance, greatest first. (Circle the right answer).
The resistances are: R1 = ρ
(i)
L
A
R2 = ρ
1.5L
= 3R1
A/2
R3 = ρ
L/2
= R1
A/2
→
R2 > R1 = R3
R2 = R3 > R 1
[(ii)] R2 > R1 = R3
(iii)
All tie
(b) (4 pts) Rank the cylinders according to the current through them, greatest first, when the same
potential difference V is placed across their lengths. (Circle the right answer).
The current is:
i=
V
R
and V1 = V2 = V3
therefore i2 < i1 = i3
[(i)] i1 = i3 > i2
(ii)
i2 > i1 = i3
(iii) All tie
4
Problem 2 [17 points]
The circuit in figure below is driven by a battery with an emf of E = 12 V and an internal resistance of
r = 1.0 Ω. In addition, there are three resistors R1 = 9.0 Ω, R2 = 4.0 Ω, and R3 = 4.0 Ω in the circuit.
a
s
R1
i = i1
-
Show all the steps to your final answers!
i2 i3
¾
-
r
R2
6 +
Ed
−
R3
The resistors R2 and R3 are connected in parallel,
1
R2 R3
therefore: R23 = 1
=
1
R2 + R3
+ R−3
R2
The resistors r, R1 , and R23 are connected in series,
therefore: Req = r + R1 + R23
s
b
(a) (5 pts) Calculate the equivalent resistance Req of the whole circuit (all four resistances).
Req = Req = r + R1 + R23 = r + R1 +
R2 R3
(4)(4)
= (1 + 9 +
)Ω = 12 Ω
R2 + R3
8
(b) (6 pts) What is the rate at which energy is dissipated in R2 ?
Energy dissipated across R2 is:
P2 = i2 R22 =
V22
R2
One way to solve it:
E
12 V
=
= 1A
Req
12 Ω
The voltage across the R2 or the R3 is, using loop rule starting from b:
The current through the source (the main current) is:
i=
+E − ir − iR1 − V2 = 0 → V2 = V3 = E − i(r + R1 ) = 12 V − (1 A)(1 + 9)(Ω) = 2 V
P2 =
V22
(2 V)2
=
= 1.0 W.
R2
4Ω
(c) (6 pts) What is the potential difference between points a and b (Va − Vb ) on the figure?
Using the loop rule, starting from the point b and only going to the point a in clockwise direction, we
have:
Vb + E − ir = Va
→
Va − Vb = E − ir = 12 V − (1 A)(1 Ω) = 11.0 V.
5
Question 3 [8 points]
Figure shows plots of potential difference V (t), across three different capacitors, as function of time
during the discharging process. All three capacitors discharge (separately) through the same resistor
R.
V
t
Vo1 = Vo3 6
C3
R
V (t) = q(t)C = qo Ce− RC
at time t = τ = RC
37%Vo1
Vo2
= 37%Vo3 p p p p p pp p pp 1
pp pp
pp p ppp p p p p p
p
p
p
p
p
37%Vo2
pp
p p
pp pp
2
pp
τ2
τ3τ1
3
- t
C2
R
C1
R
V (t) = 0.37Vo
Inspecting the figure we have:
τ2 > τ1 > τ3
→ RC2 > RC1 > RC3
(a) (4 pts) Rank the plots according to the capacitances C1 , C2 , and C3 of the capacitors 1, 2, and 3,
greatest first. (Circle the right answer).
(i)
C 1 > C 3 > C2
(ii)
C 3 > C 2 > C1
[(iii)] C2 > C1 > C3
(b) (4 pts) What is the relation between the initial charge on the capacitors 1 and 3? (Circle the right
answer).
From the figure:
V01 = Vo3
Using q = CV we have:
→
C1
qo1
=
>1
qo3
C3
→
Vo1 =
qo1
C1
and Vo3 =
qo3
C3
qo1 > qo3
[(i)] q3 < q1
(ii)
q3 > q 1
(iii) q3 = q1
6
Problem 3 [17 points]
A capacitor C is connected in series with a resistor R, an ideal battery of terminal potential E, and
a switch S. Capacitor C is initially uncharged, but at time t = 0 switch S is closed and the capacitor
charges through resistor R.
S
q©
NOTE: Express all your answers in terms of the known
quantities R, C, and E.
R
E
Charging of C in the series RC circuit:
t
q(t) = CE(1 − e− RC )
C
(a) (4 pts) What is the final potential energy stored in the capacitor?
At the moment when the capacitor is fully charged (at t = ∞) it has the charge qo = CE. The
potential difference across the capacitor is Vc = E, because the current does not exist any more, so iR
= 0. Therefore,
qo2
CE 2
Uo =
=
2C
2
(b) (4 pts) What is the current through the resistor immediately after the switch is closed, io , (at
time t = 0)?
At the time t = 0, the potential difference across the capacitor is zero, so the potential difference across
the resistance iR is equal to the emf. So, the current is maximum and equal to:
io =
VR
E
=
R
R
(c) (5 pts) At what time t is the current 37% (1/e) of the current io ?
Show all the steps in obtaining your answer.
t
t
t
dq(t)
d
E
= [CE(1 − e− RC )] = e− RC = io e− RC
dt
dt
R
t
t
io
io
= io e− RC → e−1 = e− RC
Taking the `n of both sides we
At time t: i(t) = . Therefore:
e
e
t
have:
−1 = − RC → t = RC = τ
which is the length of time equal to the one time constant.
i(t) =
(d) (4 pts) Determine VR (t), the potential difference across the resistor as function of time.
t
t
t
VR (t) = i(t)R = io Re− RC = VoR e− RC = Ee− RC
7
Question 4 [8 points]
~
The figure shows four orientations, at angle θ, of a magnetic dipole µ
~ in a magnetic field B.
1n ~µ
~µ
YH
H
HH
θ
θ©©
©©
¼
2n ~µ
3n
*
©©
©©
θ
- ~
B
HHθ
HH
j
~µ
4n
~ || sinθ |
The magnitude of a torque: | ~τ |=| ~µ || B
Also:
sin θ = sin (π − θ) = − sin (π + θ) = − sin (−θ)
which have all the same magnitude.
The potential energy of a dipole:
Also:
~ | cosθ
U = − | ~µ || B
cos θ = − cos(π − θ) = − cos (π + θ) = cos (−θ)
(a) (4 pts) Rank the orientations according to the magnitude of the torque on the dipole, greatest
first. (Circle the right answer).
[(i)] All tie
(ii)
| ~τ1 |>| ~τ2 |=| ~τ3 |>| ~τ4 |
(iii) | ~τ1 |>| ~τ2 |>| ~τ3 |>| ~τ4 |
(iv)
| ~τ1 |=| ~τ2 |>| ~τ3 |=| ~τ4 |
(b) (4 pts) Rank the orientations according to the potential energy of the dipole, greatest first.
(Circle the right answer).
(i)
All tie
(ii)
U 1 = U 3 > U2 = U 4
[(iii)] U1 = U2 > U3 = U4
8
Problem 4 [17 points]
A positively charged particle, with charge q = 2 mC, enters a region with a uniform electric field with
~ = (0.4 V/m) ı̂ and a uniform magnetic field with B
~ = (0.1 T) ̂. The charged particle’s initial
E
velocity is ~
vo = (3 m/s) ̂ + (4 m/s) k̂.
z
6
q ~
n
E
-
vo
7 ~
¶
¶
- y
t¶
q
~
B
q F
n
~
E
×nF~B
(a) (4 pts) What is the electric force on the charged particle as it enters the region? Present the
answer as the magnitude and direction OR use the unit vector notation.
The unit vector notation:
~ = (2 × 10−3 C)(0.4 V/m)ı̂ = (0.8 × 10−3 N)ı̂
F~E = q E
The magnitude of the electric force is:
FE = qE = (2 × 10−3 C)(0.4 V/m) = 0.8 mN
~ which is in the positive x direction.
The direction of F~E is the same as the direction of E,
(b) (7 pts) What is the magnetic force on the charged particle as it enters the region? Present the
answer as the magnitude and direction OR use the unit vector notation.
~ = (2 × 10−3 C)(3̂ + 4k̂)(m/s) × (0.1 T)̂ = −(0.8 × 10−3 N)ı̂
The unit vector notation: F~B = q~v × B
because ̂ × ̂ = 0, and k̂ × ̂ = −ı̂
The magnitude of the magnetic force is:
FB = qvBsinθ = qvz B = (2 × 10−3 C)(4 m/s)(0.1 T) = 0.8 × 10−3 N
The direction of F~B is into the page which is in negative x direction.
(c) (6 pts) What should be the magnitude of the magnetic field for the charged particle to go
through this region undeflected?
In order for a charged particle to move undeflected through an electric and magnetic filed these fields
have to be crossed (perpendicular to each other), the electric and the magnetic forces on the particle
have to have the same magnitude, and the directions of the two forces have to be opposite. In such a
case the net force acting on the particle is zero and it moves undeflected (only the nonzero net force
can change the velocity of the particle!)
Inspecting our results in parts (a) and (b) we can see that all the conditions just stated are met,
therefore our charged particle is moving undeflected through the space with the two filds given in the
problem.
9