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Exam 2: Tuesday, March 21, 5:00-6:00 PM
Test rooms:
• Instructor
• Dr. Hale
• Dr. Kurter
• Dr. Madison
• Dr. Parris
• Mr. Upshaw
• Dr. Waddill
Sections
F, H
B, N
K, M
J, L
A, C, E, G
D
• Special Accommodations
(Contact me a.s.a.p. if you need
accommodations different than for exam 1)
Room
104 Physics
125 BCH
199 Toomey
B-10 Bertelsmeyer*
G-3 Schrenk
120 BCH
Testing Center
*new room
Exam 2 will cover chapters 24.3 to 27
(energy stored in capacitors to forces and torques on currents)
Exam Reminders
• 5 multiple choice questions, 4 worked problems
• bring a calculator (any calculator that does not
communicate with the outside world is OK)
• no external communications, any use of a cell phone,
tablet, smartwatch etc. will be considered cheating
• no headphones
• be on time, you will not be admitted after 5:15pm
Exam Reminders
• grade spreadsheets will be posted the day after the exam
• you will need your PIN to find your grade
• test preparation homework 2 is posted on course website,
will be discussed in recitation tomorrow
• problems on the test preparation home work are NOT
guaranteed to cover all topics on the exam!!!
• LEAD review session Monday from 7 to 9 pm in BCH 120
Exam 2 topics
Energy Stored in Capacitors and Electric Fields, Dielectrics
Electric Current, Resistivity and Resistance
EMF, Electric Power
Resistors in Series and Parallel, Kirchhoff’s Rules
Electrical Instruments, RC Circuits
Magnetism, Magnetic Forces, Magnetic Flux, Gauss’ Law for
Magnetism, Motion of Charged Particle in Magnetic Field
Magnetic Force on Currents, Torque on a Current Loop
Exam 2 topics
• don’t forget the Physics 1135 concepts
• look at old tests (2014 to 2016 tests are on course website)
• exam problems may come from topics not covered in
test preparation homework or test review lecture
A parallel plate capacitor with plate separation d and plate area
A is charged by connecting it across a potential difference of
ΔV0. A dielectric slab that just fills the space between the plates
is inserted between the plates while the voltage source
remains connected to the plates.
If the energy stored in the capacitor increases by a factor of 4
when the dielectric is inserted, find the dielectric constant .
A
V0
d

Before:
 A
C0 =
d
A
d
A
d
V0
1   A 
2
U0 = 
  ΔV0 
2 d 
After:
 A
C1 =
d
1    A 
2
U1 = 
ΔV
  0  = 4U0
2 d 
V0

U1 = 4U0
U1
=4
U0
1   A 
2
ΔV

 0 
2 d 
=4
1   A 
2

  ΔV0 
2 d 
=4
For the system of resistors shown below R1=2, R2=3,
R3=6, and R4=4. If I=2A calculate
(a) the equivalent resistance,
(b) V0,
(c) the current through each resistor, and
(d) the potential difference across each resistor.
R1=2
I=2
V0
R2=3
R4=4
R3
=6
(a) Calculate the equivalent resistance.
R2 and R3 are in parallel.
R1=2
I=2
1
1
1 1 1 3 1
=
+ = + = =
R 23 R 2 R 3 3 6 6 2
R2
=3
V0
R4=4
R 23 = 2 
R1, R23, and R4 are in series.
R eq = R1 +R 23 +R 4 = 2+2+ 4 = 8 
R1=2
I=2
V0
R23=2
R4=4
R3
=6
(b) Calculate V0.
I=2
V0 = IR eq =  2 8 = 16 V
V0
Req=8
(c) Calculate the current through each resistor.
I1 = I 4 = 2 A
V23 = IR 23 =  22 = 4 V = V2 = V3
R1=2
I=2
R2
=3
V0
R4=4
V2 4
I2 =
= A
R2 3
V3 4 2
I3 =
= = A
R3 6 3
Check: I2 + I3 = 4/3 + 2/3 = 2 A, as it must be.
R3
=6
(d) Calculate the potential difference across each resistor.
V1 = IR1 = (2)(2) = 4 V
V2 = V3 = 4 V
R1=2
I=2
R2
=3
V0
Calculated in part (c).
V4 = IR4 = (2)(4) = 8 V
Check: V1 + V23 + V4 = 4 + 4 + 8 = 16 V.
Agrees with part (c).
R4=4
R3
=6
In the circuit shown R1=5.0 k, R2=10 k, and V = 12.0 V.
The capacitor is initially uncharged. After the switch has been
closed for 1.30 μs the potential difference across the capacitor
is 5.0 V. (a) Calculate the time constant of the circuit. (b) Find
the value of C. (c) Sketch the current, charge, and potential
difference across the capacitor as a function of time.
Someday, when I
have time, I will
make this into a
nice diagram!
2.41x10-6 s
0.16 nF
skip to slide 24
In the circuit shown R1=5.0 k, R2=10 k, and V = 12.0 V.
The capacitor is initially uncharged. After the switch has been
closed for 1.30 μs the potential difference across the capacitor
is 5.0 V. (a) Calculate the time constant of the circuit..
We can’t use  = RC because we don’t know C.
We are told the capacitor is charging, and given information
about potential difference, so we derive an equation for V(t).
t


RC
q  t  = Qfinal 1- e 


t


RC
C V t  = C V0 1- e 


t


RC
V t  = V0 1- e 


In the circuit shown R1=5.0 k, R2=10 k, and V = 12.0 V.
The capacitor is initially uncharged. After the switch has been
closed for 1.30 μs the potential difference across the capacitor
is 5.0 V. (a) Calculate the time constant of the circuit.
Let T = 1.30x10-6 s (for simplicity of writing equations).
T


RC
V T  = V0 1- e 


T

V T  
RC
= 1- e 
V0


e
-
T
RC
 V T  
= 1
V
0 

In the circuit shown R1=5.0 k, R2=10 k, and V = 12.0 V.
The capacitor is initially uncharged. After the switch has been
closed for 1.30 μs the potential difference across the capacitor
is 5.0 V. (a) Calculate the time constant of the circuit..
 V T  
T
= ln 1
RC
V
0 

Take natural log of both sides of
last equation on previous slide.
T
= RC = 
 V T  
ln 1
V
0 

1.3 10-6
== 2.41 10-6 s
 5 
ln 1- 
 12 
In the circuit shown R1=5.0 k, R2=10 k, and V = 12.0 V.
The capacitor is initially uncharged. After the switch has been
closed for 1.30 μs the potential difference across the capacitor
is 5.0 V. (b) Find the value of C.
 = 2.41 10-6 s = RC


2.41 10-6
C= =
=
R R1 +R 2 
15 103
C = 0.161 10-9 F= 0.161 nF
In the circuit shown R1=5.0 k, R2=10 k, and V = 12.0 V.
The capacitor is initially uncharged. (c) Sketch the current,
charge, and potential difference across the capacitor as a
function of time.
Charging capacitor (you aren’t required to derive these
equations):
t


RC
q  t  = Qfinal 1- e 


time
Charging Capacitor
charge
0.01
q (C)
0.008
0.006
0.004
0.002
0
0
0.2
0.4
0.6
0.8
1
0.8
1
0.8
1
t (s)
current
0.04
I (A)
ε - RCt
It = e
R
Charging Capacitor
0.05
0.03
0.02
0.01
0
0
time
t (s)
0.4
0.6
Charging Capacitor
0.01
voltage
0.008
q (C)
t


RC
V t  = ε 1- e 


0.2
0.006
0.004
0.002
0
0
0.2
time
t (s)
0.4
0.6
In the circuit shown R1=5.0 k, R2=10 k, and V = 12.0 V.
The capacitor is allowed to fully charge and the battery
removed from the circuit. How long does it take for the voltage
across the capacitor to drop to ¼ of its fully-charged value?
Capacitor has been charged to Qfinal = CV0 and is now
discharging.
q(t) = Q0 e
-
t
RC
CV(t) = CV0 e
V(t) = V0 e
t
V(t)
= e RC
V0
-
t
RC
-
t
RC
In the circuit shown R1=5.0 k, R2=10 k, and V = 12.0 V.
The capacitor is allowed to fully charge and the battery
removed from the circuit. How long does it take for the voltage
across the capacitor to drop to ¼ of its fully-charged value?
 V(t) 
t
ln 
=RC
 V0 
Take natural log of both sides of
last equation on previous slide.
 V(t) 
 V(t) 
t = -RC ln 
 = -  ln 

V
V
 0 
 0 
 V0
t = - ln  4
 V0


 = - 2.41 10-6 ln  1 

  4 


t = 3.34 10-6 = 3.34 s
A velocity selector consists of magnetic and electric fields. The
magnetic field is described by the expression B  Bjˆ . If B =
0.015 T find the magnitude and direction of E such that an
electron with 750 eV kinetic energy moves along the positive xaxis undeflected.
B
y
z
FB = qv  B =  -e  viˆ  Bjˆ = -evBkˆ
 FB
FE
E
v
x
FB = -e vB = evB
FB =FE
 evB = eE
FE must be in the +z direction,
so E must be in the –z
direction (because the charge
on the electron is negative.
FE = qE = -e E = eE
A velocity selector consists of magnetic and electric fields. The
magnetic field is described by the expression B  Bjˆ . If B =
0.015 T find the magnitude and direction of E such that an
electron with 750 eV kinetic energy moves along the positive xaxis undeflected.
evB =eE
B
y
z
x
 FB
FE
E
v
E = vB
1
2K
2
me v =K  v =
2
me
2K
E =B
me
A velocity selector consists of magnetic and electric fields. The
magnetic field is described by the expression B  Bjˆ . If B =
0.015 T find the magnitude and direction of E such that an
electron with 750 eV kinetic energy moves along the positive xaxis undeflected.
B
y
z
 FB
FE
E
x
electron mass is
on OSE sheet
v
2K
E =B
me
E =  0.015 
don’t forget to
convert eV to J
2  750 1.6 10-19 
9.11 10-31
E =  0.015  1.623 107 
5 N
E = 2.43  10
C
this is speed in
m/s; getting close
to relativistic
A velocity selector consists of magnetic and electric fields. The
magnetic field is described by the expression B  Bjˆ . If B =
0.015 T find the magnitude and direction of E such that an
electron with 750 eV kinetic energy moves along the positive xaxis undeflected.
B
y
z
 FB
FE
E
Problem asks for magnitude
and direction, so
v
N
, in the - z direction
C
Also “legal:”
x
E = 2.43 105
E = 2.43 105

5 N ˆ
E = -  2.43 10  k
C

N
, into page
C
E = 2.43 105
N
,
C