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2135 Exam?
That reminds
me… must
pick up test
prep HW.
adapted from http://www.nearingzero.net (nz118.jpg)
Announcements
 Grades spreadsheets will be posted the day after Exam 1. You
will need your PIN to find your grade. If you haven’t received
your PIN yet, it will be on your exam or it will be emailed to
you.*
If you lose your PIN, ask your recitation instructor. If you
haven’t received it yet, ask your instructor about it tomorrow in
recitation.
*No PIN = you can’t look up your exam grade ahead of time! Have to go to recitation.
 Physics 2135 Test Rooms, Spring 2016:
Instructor
Dr. Hale
Dr. Kurter
Dr. Madison
Mr. Noble
Dr. Parris
Mr. Upshaw
Sections
G, J
D, F
H, N
B, E
K, M
A, C, L
Special Accommodations
Exam is from
5:00-6:00 pm!
Room
125 BCH
104 Physics
199 Toomey
B-10 Bert.
G-31 EECH
G-3 Schrenk
Testing Center
Know the exam time!
Find your room ahead of time!
If at 5:00 on test day you are lost, go to 104 Physics and check the exam
room schedule, then go to the appropriate room and take the exam there.
 Reminders:
No external communication allowed while you are in the
exam room. No texting! No cell phones!
Be sure to bring a calculator! You will need it.
No headphones.
Do you know the abbreviations for:
millimicronanopico-?
Will you know them by 5:00 pm tomorrow?
Exam 1
Review
The fine print: the problems in this lecture are the standard “exam review
lecture” problems and are not a guarantee of the exam content.
Please Look at Prior Tests!
Caution: spring 2011 exam 1 did not cover capacitors—not true this semester.
Overview
Electric charge and electric force
Coulomb’s Law
Electric field
calculating electric field
motion of a charged particle in an electric field
Gauss’ Law
electric flux
calculating electric field using Gaussian surfaces
properties of conductors
Exam problems may come from topics not covered on
test preparation homework or during the review lecture.
Overview
Electric potential and electric potential energy
calculating potentials and potential energy
calculating fields from potentials
equipotentials
potentials and fields near conductors
Capacitors
capacitance of parallel plates, concentric cylinders,
(concentric spheres not for this exam)
equivalent capacitance of capacitor network
Don’t forget concepts from physics 1135 that we used!
Exam problems may come from topics not covered on
test preparation homework or during the review lecture.
If you need to evaluate an integral on tomorrow’s exam, you
will be given the integral. Exception: xn where n is a real
number.
If you don’t know what you are doing, pretend that you do and
write stuff down. You might know more than you think you do!
A positive point charge Q1 = +Q is located at position (0,D)
and a negative point charge Q2 = -2Q is located at position
(L,D). What is the electric field at the point (x=L, y=0)?
Express your answer in unit vector notation using the
symbols given above and the constants k or 0.
y
L
Q1=+Q
D
Q2=-2Q
Q
Ek 2
r
D
x
What is the electric field at the point (x=L, y=0)?
E  E1  E 2   E1x  ˆi   E1y  E 2y  ˆj


E1x  E1 cos    k


y

Q

2 
2
2
L  D  

 

L
L2  D2


kQL


2
2 32
 L  D 

L
Q1=+Q
Q2=-2Q
D
E2
E1y 
D

E1

x
E2y
L
2
kQD
D
2kQ
 2
D

2 32
What is the electric field at the point (x=L, y=0)?

 

kQL
kQD
2kQ  ˆ
ˆ



E
i 
 2 j
3
2
3
2
  L2  D2     L2  D2 
D 

 

y
You could factor kQ out of the
expression on the right hand side,
but I don’t see that it simplifies
anything much, so let’s put a box
around our answer and call it done.
L
Q1=+Q
Q2=-2Q
D
D
E2

E1

x
A negative point charge -q is placed at (x=L, y=0). What is the
electric force on the point charge? Express your answer in unit
vector notation using the symbols above and the constants k
or 0.
y
L
Q1=+Q
Let’s be smart here…
Q2=-2Q
F  qE
D
D
-q

F   q  E from previous slide
x


 
 
kQL
kQD
2kQ  ˆ 

ˆ



F   q 
i 
 2 j
3
2
3
2
2
2
  L2  D2




D
L

D
   

 
 
Again, you could factor kQ out of the expression on the right hand side, or
multiply both terms by –q. How about instead let’s put a box around our
answer and call it done.
y
L
Q1=+Q
Q2=-2Q
D
D
-q
x
An insulating spherical shell has an inner radius b and outer radius c. The
shell has a uniformly distributed total charge +Q. Concentric with the shell
is a solid conducting sphere of total charge +2Q and radius a<b. Find the
magnitude of the electric field for r<a.
Use first and last slide for
in-person lecture; delete for
video lecture
This looks like a test preparation homework problem, but it is different!
An insulating spherical shell has an inner radius b and outer radius c. The
shell has a uniformly distributed total charge +Q. Concentric with the shell
is a solid conducting sphere of total charge +2Q and radius a<b. Find the
magnitude of the electric field for r<a.
+Q
For 0<r<a, we are inside
the conductor, so E=0.
If E=0 there is no need to
specify a direction (and the
problem doesn’t ask for one
anyway).
b
a
c
+2Q
An insulating spherical shell has an inner radius b and outer radius c. The
shell has a uniformly distributed total charge +Q. Concentric with the shell
is a solid conducting sphere of total charge +2Q and radius a<b. Use Gauss’
Law to find the magnitude of the electric field for a<r<b.
q enclosed
 E  dA  o
2Q
E  4r  
o
2
+Q
b
c
a
r
+2Q
Q
E
2 o r 2
Be able to do this: begin with a statement of Gauss’s Law. Draw an
appropriate Gaussian surface on the diagram and label its radius r. Justify
the steps leading to your answer.
An insulating spherical shell has an inner radius b and outer radius c. The
shell has a uniformly distributed total charge +Q. Concentric with the shell
is a solid conducting sphere of total charge +2Q and radius a<b. Use Gauss’
Law to find the magnitude of the electric field for b<r<c.
q enclosed
 E  dA  o
E  4r
2

+Q
qshell,enclosed  q conductor,enclosed
o
b
c
q conductor,enclosed  2Q
q shell,enclosed  shell Vshell,enclosed
Qshell

Vshell,enclosed
Vshell
+2Q
a
r
q shell,enclosed
qshell,enclosed
Qshell

Vshell,enclosed
Vshell
Q

+2Q
a
r
c
Q  r 3  b3 
c
3
 b3 
Q  r 3  b3 
E  4r
b
4 3 4 3

r  b 

3
4 3 4 3 3


c


b
 3

3
q shell,enclosed 
2
+Q
c
3
b
3

o
 2Q
The direction of E is shown
in the diagram. Solving for
the magnitude E (do it!) is
“just” math.
Q  r 3  b3 
E  4r
2

c
3
b
3

o
+Q
 2Q
b
+2Q
a
r
c
What would be different if we had concentric cylinders instead of concentric spheres?
What would be different if the outer shell were a conductor instead of an insulator?
An insulating spherical shell has an inner radius b and outer radius c. The
shell has a uniformly distributed total charge +Q. Concentric with the shell
is a solid conducting sphere of total charge +2Q and radius a<b. Find the
magnitude of the electric field for b<r<c.
Qshell
4 3 4 3
   c  b 
3
3

q enclosed
 E  dA  o
4 3 4 3
  r  b   2Q
3
3


2
E  4r  
o
What would be different if we had concentric cylinders instead of concentric spheres?
What would be different if the outer shell were a conductor instead of an insulator?
Two equal positive charges Q are located at the base of an
equilateral triangle with sides of length a. What is the potential
at point P (see diagram)?
q
V=k
r
P
Q
Q
Q
VP = k  k = 2k
a
a
a
a
Q
Q
What would you do differently if
you were told Q is negative?
“No way you are going to
give us a test problem this
easy!”
Are you going to complain if
we do?
An electron is released from rest at point P. What path will the
electron follow? What will its speed be when it passes closest
to either charge Q?
P
-
a
Q
Q
An electron is released from rest at point P. What path will the
electron follow? What will its speed be when it passes closest
to either charge Q?
0
Ef  Ei   Wother if
vi=0 - initial
0
K f  U f  K i  Ui  0
a
Q
a/2
K f  Uf  Ui =  U =  q V
v
a/2
final
Q
Don’t mix up your big V’s and little v’s!
1
mv 2    e   Vf  Vi 
2


1
Q
Q
mv 2  e  2k
 2k 
2
a
 a

2

An electron is released from rest at point P. What path will the
electron follow? What will its speed be when it passes closest
to either charge Q?


1
Q
Q
mv 2  e  2k
 2k 
2
a
 a

2

vi=0 - initial
1
Q
 Q
2
mv  e  4k  2k 
2
a
a

a
Q
a/2
v
a/2
final
Q
1
 Q
2
mv  e  2k 
2
 a
Qe
v2 k
ma
For the capacitor system shown, C1=6.0 F, C2=2.0 F, and
C3=10.0 F. (a) Find the equivalent capacitance.
C1=6F
V0
C2=2F
C3=10F
C23 = C2 + C3 = 2 + 10 = 12μF
For the capacitor system shown, C1=6.0 F, C2=2.0 F, and
C3=10.0 F. (a) Find the equivalent capacitance.
C1=6F
V0
C23=12F
1
1
1
1 1
2
1
3
1
=
+
= +
=
+
=
=
Ceq
C1 C23
6 12
12 12
12
4
C eq = 4μF
Don’t expect the equivalent capacitance to always be an
integer!
For the capacitor system shown, C1=6.0 F, C2=2.0 F, and
C3=10.0 F. (b) The charge on capacitor C3 is found to be 30.0
C. Find V0.
C1=6F
C2=2F
V0
3=10F
C3C=10F
Q3= 30C
V3= ?
There are several correct
ways to solve this. Shown
here is just one.
Q = CV
V =
Q
C
V3 = V2 = V23 =
Q3
30
=
= 3V
C3
10
For the capacitor system shown, C1=6.0 F, C2=2.0 F, and
C3=10.0 F. (b) The charge on capacitor C3 is found to be 30.0
C. Find V0.
C1=6F
C23=12F
Q23= ?
V23= 3V
V0
Q23 = C23 V23 = 12 3 = 36 μC = Q1 = Qeq = Ceq V0
V0 =
36
36
=
= 9V
Ceq
4