Physics

2135

Exam

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

That reminds

me… must pick up test prep HW.

2135 Exam?

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.

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.

Exam is from 5:00-6:00 pm!

Physics 2135 Test Rooms, Fall 2015:

Instructor Sections RoomDr. Hagen J, R G-3 SchrenkDr. Hale K, P 120 Butler-CarltonDr. Hor B, D G-3 SchrenkDr. Kurter A, C 104 PhysicsDr. Madison G, L 204 McNuttDr. Parris M, Q 125 Butler-CarltonMr. Upshaw N, H G-5 Humanities/Soc. Sci.Dr. Vojta E, F 199 Toomey

Special Accommodations Testing Center

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.

*Fall 2015: there are a handful of very simple numerical calculations which you should be able to do without a calculator, but you may bring one anyway just in case you feel a need to use it.

Do you know the abbreviations for:

milli-

micro-

nano-

pico-?

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 forceCoulomb’s Law

Electric fieldcalculating electric fieldmotion of a charged particle in an electric field

Gauss’ Lawelectric fluxcalculating electric field using Gaussian surfacesproperties of conductors

Exam problems may come from topics not covered ontest preparation homework or during the review lecture.

OverviewElectric potential and electric potential energy

calculating potentials and potential energycalculating fields from potentialsequipotentialspotentials and fields near conductors

Capacitorscapacitance 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 ontest 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, and sine & cosine.

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!

L

x

y

P

y

An infinitesimal length dx of rod has dq=dx of charge, where =Q/L.

Ex from a dq of charge with x=|x1|>0 will be equal and magnitude and opposite in direction to Ex from a dq of charge with x=-|x1|<0.Thus, by symmetry, Ex=0.

A thin rod of length L with total charge Q lies along the x-axis as shown. Find the magnitude and direction of the electric field at P, a distance y from the rod and along the perpendicular bisector of the rod.

dqdq

L

x

y

P

y

dq

dE

y 2 2

kdq k dxdE dE cos cos cos

r r

r

ycos

r 2 2r x y

y yE dEL/2

2 2 2 2L/2

dx yk

x y x y

Instead of completely working the above problem in the live lecture, I reminded students that this is the same as problem 21.52, which you had for homework.

On the exam, you will be able to evaluate all integrals without needing a table of integrals. Know how to integrate xn, where n is a positive or negative number.

The video lecture shows an alternative way of doing the integral, which you won’t be responsible for on Exam 1.

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

Q

dq = dslength 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

xdE

ds

dds = 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

4x 0

E = - dE cos

12 =

8 4

4y 0

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 4x 2 20 0

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

a a

4x 2 20

4k Q 4k QE = sin = sin sin 04a 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 4y 2 20 0

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

a a

4y 2 20

4k Q 4k QE = cos = cos cos 04a 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.

a

y

x

What would you do differently if we placed a second eighth of a circle in the fourth

quadrant, as shown?

What would be different if the charge were negative?

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!

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

2o

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 3shell,enclosed

3 3

Q 4 4q r b

4 4 3 3c b3 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 3shell

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?

QQ

P

a

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)?

qV = k

r

P

Q Q QV = k k = 2k

a a a

“No way you are going to give us a test problem this easy!”

Are you going to complain if we do?

What would you do differently if you were told Q is negative?

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?

QQ

P

a

-

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?

QQ

a

-

-v

vi=0 initial

final

f i other i fE E W

f f i iK U K U 0

f f iK U U = U = q V

2f i

1mv e V V

2 a/2 a/2

21 Q Qmv e 2k 2k

a2 a2

Don’t mix up your big V’s and little

v’s!

0

0

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?

QQ

a

-

-v

vi=0 initial

final

a/2 a/2

21 Q Qmv e 2k 2k

a2 a2

21 Q Qmv e 4k 2k

2 a a

21 Qmv e 2k

2 a

Qev 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=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=10FC3=10FQ3= 30C

V3= ?

Q = CV

QV =

C3

3 2 233

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=12FQ23= ?

V23= 3V

0eq

36 36V = = = 9 V

C 4