Exam 2: Tuesday, March 21, 5:-6: PM Test rooms: Instructor Sections Room Dr. Hale F, H 14 Physics Dr. Kurter B, N 125 BCH Dr. Madison K, M 199 Toomey Dr. Parris J, L B-1 Bertelsmeyer* Mr. Upshaw A, C, E, G G-3 Schrenk Dr. Waddill D 12 BCH Special Accommodations (Contact me a.s.a.p. if you need accommodations different than for exam 1) Testing Center *new room Exam 2 will cover chapters 24.3 to 27 (energy stored in capacitors to forces and torques on currents)
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 test preparation homework 2 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!!!
Exam 2 topics Energy Stored in Capacitors and Electric Fields, Dielectrics Electric Current, Resistivity and Resistance EMF, Electric Power Resistors in Series and Parallel, Kirchhoff s Rules Electrical Instruments, RC Circuits Magnetism, Magnetic Forces, Magnetic Flux, Gauss Law for Magnetism, Motion of Charged Particle in Magnetic Field Magnetic Force on Currents, Torque on a Current Loop
Exam 2 topics don t forget the Physics 1135 concepts look at old tests (214 to 216 tests are on course website) exam problems may come from topics not covered in test preparation homework or test review lecture
A parallel plate capacitor with plate separation d and plate area A is charged by connecting it across a potential difference of ΔV. 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. A d V
Before: C = d A A d 1 A ΔV U = 2 d 2 V After: C = 1 A d 1 A 2 U= ΔV = 4U 2 d 1 V A d
U 1= 4U U 1 =4 U 1 A 2 d 1 A 2 d ΔV ΔV 2 2 =4 =4
For the system of resistors shown below R 1 =2, R 2 =3, R 3 =6, and R 4 =4. If I=2A calculate (a) the equivalent resistance, (b) V, (c) the current through each resistor, and (d) the potential difference across each resistor. I=2 R 1 =2 V R 4 =4 R 2 =3 R 3 =6
(a) Calculate the equivalent resistance. R 2 and R 3 are in parallel. I=2 R 1 =2 1 1 1 1 1 3 1 = + = + = = R R R 3 6 6 2 23 2 3 V R 4 =4 R 2 =3 R 3 =6 R 23 = 2 R 1, R 23, and R 4 are in series. I=2 R 1 =2 R eq = R 1 +R 23 +R 4 = 2+2+4 = 8 V R 4 =4 R 23 =2
(b) Calculate V. V = IR eq = 2 8 = 16 V R eq =8 I=2 V
(c) Calculate the current through each resistor. I 1 = I 4 = 2 A I=2 R 1 =2 V 23 = IR 23 = 22 = 4 V = V 2 = V3 V R 4 =4 R 2 =3 R 3 =6 V 4 2 I 2 = = A R2 3 V 4 2 3 I 3 = = = A R3 6 3 Check: I 2 + I 3 = 4/3 + 2/3 = 2 A, as it must be.
(d) Calculate the potential difference across each resistor. V 1 = IR 1 = (2)(2) = 4 V I=2 R 1 =2 V 2= V 3= 4 V Calculated in part (c). V R 4 =4 R 2 =3 R 3 =6 V 4 = IR 4 = (2)(4) = 8 V Check: V 1 + V 23 + V 4 = 4 + 4 + 8 = 16 V. Agrees with part (c).
In the circuit shown R 1 =5. k, R 2 =1 k, and V = 12. V. The capacitor is initially uncharged. After the switch has been closed for 1.3 μs the potential difference across the capacitor is 5. V. (a) Calculate the time constant of the circuit. (b) Find the value of C. (c) Sketch the current, charge, and potential difference across the capacitor as a function of time. Someday, when I have time, I will make this into a nice diagram! 2.41x1-6 s.16 nf skip to slide 24
In the circuit shown R 1 =5. k, R 2 =1 k, and V = 12. V. The capacitor is initially uncharged. After the switch has been closed for 1.3 μs the potential difference across the capacitor is 5. 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). q t = Q final C V t = C V 1- e t - RC 1- e t - RC V t = V 1- e t - RC
In the circuit shown R 1 =5. k, R 2 =1 k, and V = 12. V. The capacitor is initially uncharged. After the switch has been closed for 1.3 μs the potential difference across the capacitor is 5. V. (a) Calculate the time constant of the circuit. Let T = 1.3x1-6 s (for simplicity of writing equations). V T = V 1- e T - RC T VT = 1- e V - RC T - e RC = 1- VT V
In the circuit shown R 1 =5. k, R 2 =1 k, and V = 12. V. The capacitor is initially uncharged. After the switch has been closed for 1.3 μs the potential difference across the capacitor is 5. V. (a) Calculate the time constant of the circuit.. T - =ln1- RC VT V Take natural log of both sides of last equation on previous slide. T - =RC = VT ln1- V -6 1.31-6 = - = 2.411 s 5 ln1-12
In the circuit shown R 1 =5. k, R 2 =1 k, and V = 12. V. The capacitor is initially uncharged. After the switch has been closed for 1.3 μs the potential difference across the capacitor is 5. V. (b) Find the value of C. -6 = 2.411 s = RC 2.411 C = = = 3 R R +R 151 1 2-6 -9 C =.1611 F=.161 nf
voltage current charge In the circuit shown R 1 =5. k, R 2 =1 k, and V = 12. V. The capacitor is initially uncharged. (c) Sketch the current, charge, and potential difference across the capacitor as a function of time. Charging capacitor (you aren t required to derive these Charging Capacitor equations): t - qt = Q RC final 1- e time t (s) q (C).1.8.6.4.2.2.4.6.8 1 t ε It = e - RC R I (A) Charging Capacitor.5.4.3.2.1.2.4.6.8 1 time t (s) ε t V t = 1- e - RC q (C).1.8.6.4.2 Charging Capacitor.2.4.6.8 1 time t (s)
In the circuit shown R 1 =5. k, R 2 =1 k, and V = 12. 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 Q final = CV and is now discharging. q(t) = Q e t - RC CV(t) = CV e V(t) = V e t - RC t - RC V(t) = e V t - RC
In the circuit shown R 1 =5. k, R 2 =1 k, and V = 12. 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? V(t) ln = - V t RC Take natural log of both sides of last equation on previous slide. V(t) V(t) t = -RC ln = - ln V V V -6 1 t = - ln 4 = - 2.411 ln V 4 t = 3.34 1-6 = 3.34 s
A velocity selector consists of magnetic and electric fields. The magnetic field is described by the expression B Bj ˆ. If B =.15 T find the magnitude and direction of E such that an electron with 75 ev kinetic energy moves along the positive x- axis undeflected. z y x B F B - v F E E ˆ ˆ ˆ F = qv B = -e vi Bj=-evBk B F E must be in the +z direction, so E must be in the z direction (because the charge on the electron is negative. F B = -e vb = evb F E = qe = -e E = ee F =F B E evb = ee
A velocity selector consists of magnetic and electric fields. The magnetic field is described by the expression B Bj ˆ. If B =.15 T find the magnitude and direction of E such that an electron with 75 ev kinetic energy moves along the positive x- axis undeflected. z y x B F B - v F E E evb = ee E = vb 1 2 2K mev =K v = 2 m E =B 2K m e e
A velocity selector consists of magnetic and electric fields. The magnetic field is described by the expression B Bj ˆ. If B =.15 T find the magnitude and direction of E such that an electron with 75 ev kinetic energy moves along the positive x- axis undeflected. y B F B - v F E E E =B 2K m e E =.15 don t forget to convert ev to J 2 751.6 1 9.111-31 -19 z x E =.151.623 1 7 electron mass is on OSE sheet E = 2.431 C 5 N this is speed in m/s; getting close to relativistic
A velocity selector consists of magnetic and electric fields. The magnetic field is described by the expression B Bj ˆ. If B =.15 T find the magnitude and direction of E such that an electron with 75 ev kinetic energy moves along the positive x- axis undeflected. z y B x F B - v F E E Problem asks for magnitude and direction, so E = 2.431 5 N Also legal: C E = -2.431, 5 N in the - z direction C k ˆ E = 2.431, C 5 N into page E = 2.431 C 5 N,