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The Physics 2135 graders are ready to begin

grading.

Pick up test prep HW!

Exam 2

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. The contents of this lecture may change at the

last minute!

Please Look at Prior Tests!

Announcements

Pick up test preparation homework from the back of the lecture hall.

You will need a calculator. You may not use cell phones, etc. as calculators.

Updated course grades should be available by 3:00 pm on the day after the exam. Grades spreadsheets will not be updated over the weekends after exams.

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

Look at posted prior tests to get an idea of type and difficulty of exam 2 problems.

Please Look at Prior Tests!

If there are any problems with posted exams or solutions, let me know before exam day, so you have a chance to see corrections!

Reminders:

No external communication allowed while you are in the exam room.

Please do not leave the exam room before 5:30 pm.

No headphones.

Be sure to bring a calculator!

What’s The Big Idea?

Helpful Hints

You will not be required to do a Kirchhoff’s rules problem in which you must solve three or more simultaneous equations… but you must be able to apply

V 0 around any closed circuit loopI 0 at any circuit junction

How to determine Afor a current loop.

Curl your fingers along the loop in the direction of the (conventional) current.

Your outstretched thumb points in the direction of A.

By the way, this is the consequence if we catch you using your left hand for the right hand rule…

***March, 2014: not necessary because there will be no torque on a magnetic dipole problems.

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 .

V0

A

dSkip to slide 14.

V0

A

d

V0

A d

Before:

0

AC =

d

2

0 0

A1U = ΔV

2 d

After:

1

AC =

d

2

1 0 0

A1U = ΔV =4U

2 d

1 0U =4U

=4

1

0

U=4

U

2

0

2

0

A1ΔV

2 d =4A1

ΔV2 d

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

R2=3 R3

=6

R4=4

V0

I=2

Skip to slide 19.

(a) Calculate the equivalent resistance.

R1=2

R2

=3R3

=6

R4=4

V0

I=2

R1=2

R4=4

V0

I=2

R2 and R3 are in parallel.

323 2

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

R R R 3 6 6 2

23R = 2

R23=2

R1, R23, and R4 are in series.

eq 1 23 4R = R +R +R =2+2+4=8

(b) Calculate V0.

V0

I=2

0 eqV = IR = 2 8 = 16 V Req=8

(c) Calculate the current through each resistor.

R1=2

R2

=3R3

=6

R4=4

V0

I=2I1 = I4 = 2 A

22

2

V 4I = = A

R 3

Check: I2 + I3 = 4/3 + 2/3 = 2 A, as it must be.

23 23 2 3V = IR = 2 2 = 4 V= V = V

33

3

V 4 2I = = = A

R 6 3

(d) Calculate the potential difference across each resistor.

R1=2

R2

=3R3

=6

R4=4

V0

I=2V1 = IR1 = (2)(2) = 4 V

Check: V1 + V23 + V4 = 4 + 4 + 8 = 16 V.Agrees with part (c).

2 3V = V = 4 V Calculated in part (c).

V4 = IR4 = (2)(4) = 8 V

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.

2.41x10-6 s

0.16 nF

Someday, when I have time, I will make this into a nice diagram!

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

finalq t =Q 1-e

t-RC

0C V t =C V 1-e

t-RC

0V t =V 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.

T-RC

0V T =V 1-e

Let T = 1.30x10-6 s (for simplicity of writing equations).

T-RC

0

V T= 1-e

V

T-RC

0

V Te = 1-

V

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..

0

V TT- =ln 1-RC V

0

T- =RC=

V Tln 1-

V

-6-61.3 10

=- =2.41 10 s5

ln 1-12

Take natural log of both sides of last equation on previous slide.

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.

-6=2.41 10 s = RC

-6

31 2

2.41 10C= = =

R R +R 15 10

-9C=0.161 10 F=0.161 nF

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.

t-RC

0q(t)=Q e

t-RC

0CV(t)=CV e

t-RC

0V(t)=V e

t-RC

0

V(t)= e

V

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?

0

V(t) tln = -

V RC

0 0

V(t) V(t)t=-RC ln = - ln

V V

0-6

0

V14t=- ln = - 2.41 10 ln

V 4

Take natural log of both sides of last equation on previous slide.

-6t=3.34 10 = 3.34 s

Two charged particles having identical masses and velocities enter a region of uniform magnetic field, as shown in the figure. Particle has a charge Q1 and moves in a circular path of radius r1=R. Particle has a charge Q2 and moves in a circular path of radius r2=2R. The two paths are shown in the figure. What is the ratio Q2/Q1?

Q1, m, v

Q2, m, v B

2R

R

Q1, m, v

Q2, m, v B

2R

R

2 21

1 1

v vF = ma = m =m

R R

1 1 1 1 1F = Q v ×B = Q vB sin90° = Q vB

2

1

vQ vB=m

R

1

mvQ =

BRF1

Q1, m, v

Q2, m, v B

2R

R

2 2 2

2 2

v vF = ma = m =m

2R 2R

2 2 2 2 2F = Q v ×B = Q vB sin90° = Q vB

2

2

vQ vB=m

2R

2

mvQ =

2BR

F2

Q1, m, v

Q2, m, v B

2R

R

1

mvQ =

BR

2

mvQ =

2BR

2

1

mvQ BR 12BR= = =

mvQ 2BR 2BR

2

1

Q 1=-

Q 2

Q1, m, v

Q2, m, v B

2R

R

F2

F1

v1

v2

The forces are not drawn to scale: F2 =

F1/2.

Apply the right hand rule. Q1 must be positive for the F1 to be , and Q2 must be negative for F2 to be .