That reminds
me… must
download the
test prep HW.
adapted from http://www.nearingzero.net (nz118.jpg)
Exam 1: Tuesday, Feb 14, 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
Room
104 Physics
125 BCH
199 Toomey
St Pat’s Ballroom*
G-3 Schrenk
120 BCH
Testing Center
*exam 1 only
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.
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
(PINs were emailed by the recitation instructors)
• test preparation homework 1 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 1 topics
Electric charge and electric force, Coulomb’s Law
Electric field (calculating electric fields, motion of a charged
particle in an electric field, dipoles)
Gauss’ Law (electric flux, calculating electric fields via
Gaussian surfaces, fields and surface charges of conductors)
Electric potential and potential energy (calculating work,
potential energy and potential, calculating fields from potentials,
equipotentials, potentials of conductors)
Capacitors (calculating capacitance, equivalent capacitance of
capacitor network, charges and voltages in capacitor network)
Exam 1 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
Three charges +Q, +Q, and –Q, are located at the corners of
an equilateral triangle with sides of length a. What is the force
on the charge located at point P (see diagram)?
y
P
+Q
F1
F   F1x  F2x  ˆi   F1y  F2y  ˆj


F   F1 cos   F2 cos   ˆi
  F1 sin   F2 sin   ˆj
F2
a
F1  k
+Q
=60
-Q
x
F2  k
 Q  Q 
a2
 Q Q
a2
Q2
k 2
a
Q2
k 2
a
Note: if there is not a problem like this on Exam 1, there will be one on the Final!
Three charges +Q, +Q, and –Q, are located at the corners of
an equilateral triangle with sides of length a. What is the force
on the charge located at point P (see diagram)?
y
F1
P
+Q


 kQ 2
kQ 2
F   2 cos 60  2 cos 60
a
 a
ˆ
i

 kQ 2
kQ 2
  2 sin 60  2 sin 60
a
 a
F2
ˆ
j

a
+Q
-Q
x
I could have stated that Fy=0 and Fx=2F1x by
symmetry, but I decided to do the full calculation here.
Three charges +Q, +Q, and –Q, are located at the corners of
an equilateral triangle with sides of length a. What is the force
on the charge located at point P (see diagram)?
y
kQ2
F  2 2 cos 60 ˆi
a
F1
P
+Q


F
F2
kQ 2 ˆ
F 2 i
a
a
+Q
-Q
x
What is the electric field at P due to the two charges at the
base of the triangle?
You can “repeat” the above calculation,
replacing F by E (and using Coulomb’s Law).
y
Or you can be smart… F  qE
P
kQ 2
î
2
F
kQ ˆ
a
E 
 2 i
q
Q
a
a
+Q
Caution: never write q  E
-Q
F
. Why?
x
This is the charge which had
been at point P, “feeling” the
force F .
A rod is bent into an eighth of a circle of radius a, as shown.
The rod carries a total positive charge +Q uniformly distributed
over its length. What is the electric field at the origin?
y
dE = k
a
x
dq
r2
A rod is bent into an eighth of a circle of radius a, as shown.
The rod carries a total positive charge +Q uniformly distributed
over its length. What is the electric field at the origin?
y
dE = k
dq

dE

a
dq
r
2
dq
=k
dq =  ds =
a2
charge
length
ds
x
dq =
Q
 length of
arc 
ds
Q
4 Q
dq =
ds =
ds
2a
a
8


A rod is bent into an eighth of a circle of radius a, as shown.
The rod carries a total positive charge +Q uniformly distributed
over its length. What is the electric field at the origin?
y
d
ds

dE
a
ds = a d
x
A rod is bent into an eighth of a circle of radius a, as shown.
The rod carries a total positive charge +Q uniformly distributed
over its length. What is the electric field at the origin?
y
dq

dE
 4  Q 


  a d  
 a 

k
dE = 2
a

a
dE x = - dE cos 
x
dE y = - dE sin 
Ex = Ey = -


4
0


0
dE cos 
1

 2  =
8
4
4
dE sin 
A rod is bent into an eighth of a circle of radius a, as shown.
The rod carries a total positive charge +Q uniformly distributed
over its length. What is the electric field at the origin?
dE =
4k Q
Ex = 
Ex = 
2
d

4k Q
a

0
4
a
4k Q
a 2
2
cos  d = 

 sin  0
4
= 
4k Q
a
4k Q
a 2
2


0
4
cos  d
sin  4  sin 0

4k Q  2
2 2k Q
Ex = 
 0  = 

2
2
a  2

a

A rod is bent into an eighth of a circle of radius a, as shown.
The rod carries a total positive charge +Q uniformly distributed
over its length. What is the electric field at the origin?
dE =
4k Q
Ey = 
Ey = 
2
d

4k Q
a

0
4
a
4k Q
a 2
2
sin  d = 

  cos  0
4
= 
4k Q
a
2
0
4k Q
a 2
4k Q 
4k Q
2 
Ey = 
 1 = 
 
2
2
a  2

a



4
sin  d
  cos  4  cos 0

2
1 

2 

A rod is bent into an eighth of a circle of radius a, as shown.
The rod carries a total positive charge +Q uniformly distributed
over its length. What is the electric field at the origin?
 2 2kQ  ˆ  4kQ 
2 ˆ
E =  
 i   2 1 
  j
2
2 
 a   a 
2kQ 
E=  2
a 
 2  ˆi   2  2  ˆj
You should provide reasonably simplified answers
on exams, but remember, each algebra step is a
chance to make a mistake.
What would be different if the charge were negative?
What would you do differently if you were asked to
calculate the potential rather than the electric field?
How would you find the force on a test charge -q at
the origin?
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  qconductor,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?
A ring with radius R has a uniform positive charge density .
Calculate the potential difference between the point at the
center of the ring and a point on the axis of the ring that is a
distance of 3R from the center of the ring.
R
3R
Begin by deriving the equation for the potential along the
central axis of a ring of charge. We did this back in part 2 of
lecture 6. I am going to be lazy… err, efficient… and just copy
the appropriate slides.
dq
r
R
P
x
Every dq of charge on the
ring is the same distance
from the point P.
x
Q
dq
dq
dV  k
k
r
x2  R2
V
ring
dV  
ring
kdq
x2  R2

k
x2  R2

ring
dq
dq
r
R
P
x
x
Q
V
V
V
k
x R
2
2
kQ
x2  R2
2kR
x2  R2

ring
dq
Q    2R 
A ring with radius R has a uniform positive charge density .
Calculate the potential difference between the point at the
center of the ring and a point on the axis of the ring that is a
distance of 3R from the center of the ring.
R
x
V(x) 
3R
2kR
2kR
2kR
x2  R2
1 
1
V(0)  V(3R) 

 2kR  

2
2
2
2
R
R 10 

0 R
 3R   R
A ring with radius R has a uniform positive charge density .
Calculate the potential difference between the point at the
center of the ring and a point on the axis of the ring that is a
distance of 3R from the center of the ring.
R
x
3R
 10  1 
V(0)  V(3R)  2k 

 10 
If a proton is released from rest at the center of the ring,
how fast will it be at point P?
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
C3=10F
C3=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 = CeqV0
V0 =
36
36
=
= 9V
Ceq
4