adapted from http://www.nearingzero.net (nz118.jpg)

That reminds me… must

download the test prep HW.

Test rooms:

• Instructor Sections Room• Dr. Hale F, H 104 Physics• Dr. Kurter B, N 125 BCH• Dr. Madison K, M 199 Toomey• Dr. Parris J, L St Pat’s Ballroom*• Mr. Upshaw A, C, E, G G-3 Schrenk• Dr. Waddill D 120 BCH

• Special Accommodations Testing Center

*exam 1 only

Exam 1: Tuesday, Feb 14, 5:00-6:00 PM

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.

• 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 Reminders

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

F1

F2

+Q

-Q+Q

P

a

x

y

1 2

1 2

ˆF F cos F cos i

ˆ F sin F sin j

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)?

1x 2x 1y 2yˆ ˆF F F i F F j

2

1 2 2

Q Q QF k k

a a

=60

2

2 2 2

Q Q QF k k

a a

Note: if there is not a problem like this on Exam 1, there will be one on the Final!

F1

F2

+Q

-Q+Q

P

a

x

y

2 2

2 2

2 2

2 2

kQ kQ ˆF cos60 cos60 ia a

kQ kQ ˆ sin 60 sin 60 ja a

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)?

I could have stated that Fy=0 and Fx=2F1x by symmetry, but I decided to do the full calculation here.

2

2

kQ ˆF 2 cos60 ia

2

2

kQ ˆF ia

F1

F2

+Q

-Q+Q

P

a

x

y

F

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)?

-Q+Q

P

a

x

y

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).

Or you can be smart…

2

2

2

kQi

F kQa ˆE iq Q a

F qE

This is the charge which had been at point P, “feeling” the force .F

Caution: never write . Why?EqF

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?

a

y

x

2

dqdE = k

r

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?

a

y

x

2 2

dq dqdE = k = k

r a

dE

dq

chargedq = ds = ds

length

Qdq = ds

length of arc

Q 4 Q

dq = ds = ds2 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?

a

y

x

dE

ds

d

ds = a d

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?

a

y

x

2

4 QkdE = a d

a a

dE

dq

xdE = - dE cos

ydE = - dEsin

4

x0

E = - dEcos

12 =

8 4

4

y0

E = - dEsin

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

4k QdE = d

a

4 4

x 2 20 0

4k Q 4k QE = cos d = cos d

a a

4x 2 20

4k Q 4k QE = sin = sin sin 0

4a a

x 2 2

4k Q 2 2k Q2E = 0 =

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?

2

4k QdE = d

a

4 4

y 2 20 0

4k Q 4k QE = sin d = sin d

a a

4y 2 20

4k Q 4k QE = cos = cos cos0

4a a

y 2 2

4k Q 4k Q2 2E = 1 = 1

a 2 a 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 2

2 2kQ 4kQ 2ˆ ˆE = i 1 ja a 2

2

2kQ ˆ ˆE = 2 i 2 2 ja

You should provide reasonably simplified answers on exams, but remember, each algebra step is a chance to make a mistake.

What would you do differently if you were asked to calculate the potential rather than the electric field?

What would be different if the charge were negative?

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.

This looks like a test preparation homework problem, but it is different!

Use first and last slide for in-person lecture; delete for video lecture

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

c

b a

+2Q

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).

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

c

b a

+2Q

r

enclosed

o

qE dA

2

o

2QE 4 r

2

o

QE

2 r

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

c

b a+2Q

r

enclosed

o

qE dA

shell,enclosed conductor,enclosed2

o

q qE 4 r

shellshell,enclosed shell shell,enclosed shell,enclosed

shell

Qq V V

V

conductor,enclosedq 2Q

+Q

c

b a+2Q

r

shellshell,enclosed shell,enclosed

shell

Qq V

V

3 3

shell,enclosed3 3

Q 4 4q r b

4 4 3 3c b

3 3

3 3

shell,enclosed 3 3

Q r bq

c b

3 3

3 3

2

o

Q r b2Q

c bE 4 r

The direction of is shown in the diagram. Solving for the magnitude E (do it!) is “just” math.

E

+Q

c

b a+2Q

r

3 3

3 3

2

o

Q r b2Q

c bE 4 r

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?

enclosed

o

qE dA

3 3

2

o

4 4r b 2Q

3 3E 4 r

3 3

shell

4 4Q c b

3 3

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.

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.

P

R

dq

r

xx

Every dq of charge on the ring is the same distance from the point P.

dqdV k

r

2 2 2 2ring ring ring

kdq kV dV dq

x R x R

2 2

dqk

x R

Q

2 2 ring

kV dq

x R

2 2

kQV

x R

P

R

dq

r

xx

Q

Q 2 R

2 2

2 kRV

x 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

3R

2 2

2 kRV(x)

x R

2 2 2 2

2 kR 2 kR 1 1V(0) V(3R) 2 kR

R R 100 R 3R R

x

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

10 1V(0) V(3R) 2 k

10

x

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=6F

V0

C2=2F C3=10F23 3 2C = C + C = 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.

Don’t expect the equivalent capacitance to always be an integer!

C1=6F

V0

C23=12F

1 23

eq

1 1 1 1 1 2 1 3 1 = + = + = + = =

C C C 6 12 12 12 12 4

eqC = 4μF

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.

There are several correct ways to solve this. Shown here is just one.

C1=6F

V0

C2=2F C3=10F

C3=10F

Q3= 30C

V3= ?

Q = CV

QV =

C3

3 2 23

3

Q 30V = V = V = = = 3 V

C 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.

23 23 23 1 eq eq 0Q = C V = 12 3 = 36 μC = Q = Q = C V

C1=6F

V0

C23=12F

Q23= ?

V23= 3V

0

eq

36 36V = = = 9 V

C 4