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9/23/15 Capacitors with a dielectric for Fixed Amount of Charge (Q is constant) ANNOUNCEMENT *Exam 1: Monday September 28, 2015, 8 PM - 10 PM *Location: Elliot Hall of Music *Covers all readings, lectures, homework from Chapters 21 through 23. *The exam will be multiple choice (15-18 questions). For Fixed Charge, the insertion of a dielectric Increases the capacitance: decreases the Electric Potential *Practices exams can be found on the course website and on CHIP. decreases the Electric Field *Be sure to bring your student ID card and your own one-page (two-side) crib sheet. 9/23/15 Ewithdielectric = decreases the Electric Potential Energy NOTE THAT FEW EQUATIONS WILL BE GIVEN – YOU ARE REMINDED THAT IT IS YOUR RESPONSIBILITY TO CREATE WHATEVER TWO-SIDED CRIB SHEET YOU WANT TO BRING TO THIS EXAM. The equation sheet that will be given with the exam is posted on the course homepage. Click on the link on the left labeled “Equationsheet” Vwithdielectric = Cwithdielectric = KCo Vo K Eo K U withdielectric = Uo K The decrease in energy can be accounted for by the work done in moving the dielectric slab between the plates of the capacitor. 1 Capacitors with a dielectric connected to a battery (V = Constant Potential). 9/23/15 2 Chapter 25 For a constant applied potential, the insertion of a dielectric:: Cwithdielectric = KCo increases the amount of charge on the plates increases the Electric Potential Energy • • • • • Qwithdielectric = Qo K U withdielectric 2 Qwithdielectric ∝ K Current Resistivity as a property of materials Temperature dependence of resistivity Emf Power The increase in energy occurs because work is done to maintain the potential across the plates as the dielectric is inserted. 9/23/15 3 9/23/15 4 1 9/23/15 Electrostatics versus Electric Current Electric Current Electrostatic Equilibrium Steady-State + - Battery Note: Current arrow is drawn in direction in which positive charge carriers would move, even if the actual charges are negative & move in the opposite direction. 9/23/15 5 Electric Current 9/23/15 6 Drift Velocity • Notice that flow of positive charge in one direction, Becomes more negative + + + + + + + + + + Becomes more positive CURRENT I • It’s completely equivalent to negative charge in the opposite direction. Becomes Becomes more more - positive negative CURRENT I 9/23/15 7 9/23/15 8 2 9/23/15 Microscopic View of an Electron in Motion Drift Speed, Total Charge & Current # charge carriers volume q = charge of each particle vd = drift velocity n= with E without E 9/23/15 9 Relationship between Current and Drift Speed 9/23/15 Electric Current Find vd for 14-gauge copper wire carrying a current of 1 A. Assume there is 1 free electron/atom. Copper n = natoms ρN = M 10 one q in = 8.93 g/cm3 one q out M = 63.55 g/mol 14 gauge wire Area= 2.081 mm2" vd = 9/23/15 I qnA 11 9/23/15 12 3 9/23/15 Resistance Resistance & Ohm’s Law Resistance is a property of the object, i.e . It depends on the shape and material ohmic Non-ohmic 9/23/15 13 Resistivity Resistivity is a property of the material. 9/23/15 Resistivity & Temperature Coefficients area A I I 14 Material L Ag Cu W Si Si, n-type Si, p-type glass length Resistivity " (m) 1.6 x 10-8 1.7 x 10-8 5.5 x 10-8 640 8.7 x 10-4 2.8 x 10-3 1010–1014 Temp. coeff. (K-1) 3.8 x 10-3 3.9 x 10-3 4.5 x 10-3 –7.5 x 10-2 See TM Table 25-1 for more. 9/23/15 15 9/23/15 16 4 9/23/15 Temperature Dependence of Resistances ρ − ρo = ρoα (T − To ) R=ρ L A Temperature Dependence ρ = ρo (1 + α (T − To )) R = Ro (1 + α (T − To )) −3 Heated Tungsten Light Bulb filament at 3000 K: α = 4.5x10 / K for copper (Cu) as a function of temperature Notice: Resisitivity increases as temperature increases. This curve does not deviate greatly from a straight line. 9/23/15 17 DEMO: Temperature Dependence 9/23/15 18 DEMO: Temperature Dependence lamp liquid nitrogen ~ 77oK B S vacuum bottle Ge Lower Cu initially at room temperature (~ 300oK) into liquid N2. 9/23/15 Heat the Ge with a candle. semiconductor 19 9/23/15 20 5 9/23/15 Temperature Stable Resistor Power in Electric Circuits A temperature-stable resistor is to be made by connecting a resistor made of silicon in series with one made of nichrome. If the required total resistance is 1300 in a wide temperature range around 20ºC, what should be the resistances of the two resistors? Rn Rtotal = RN + RSi = 1300Ω In general: R = Ro (1 + α (T − To )) RoN (1 + α N (T − To )) + RoSi (1 + α Si (T − To )) = 1300Ω dU = dqV = idtV + RSi P= For a Resistor: P=I2R = V2/R 21 Power in Electric Circuits Units of Power: Volts Ampere= Joule/second=Watt Power associated with dissipation of U into thermal energy in the resistor. 9/23/15 22 Real Battery Take a 100 W light bulb powered by 110 Volts (RMS AC). What is the resistance of the (hot) filament? We know P and V, don’t know I or R, are asked for R. So we choose the last form below, and solve for R: Examine potential as we start from point b and end at a: internal resistance emf ε − Ir − IR = 0 ε I= R+r R = V2/P = 110 x 110/100 = 121Ω In an ideal battery: r=0 Va − Vb = ε V I= R 2 V P = IV = I 2 R = R Note that the cold filament will have ~13 times less resistance, and therefore there will be a big surge in current as the bulb is turned on. Often, used lightbulbs burn out at this instant. 9/23/15 dU = iV dt Power associated with transfer: P=IV (RoN + RoSi ) + RoN α N (T − To ) + RoSiα Si (T − To )) = 1300Ω (RoN α N + RoSiα Si )(T − To ) = 0 (RoN α N + RoSiα Si ) = 0 (1300Ω − RoSi )α N + RoSiα Si = 0 9/23/15 ? - 23 9/23/15 Note: Over the battery: Va − Vb = ε − Ir arrows always points from negative to positive. 24 6 9/23/15 Effect of Internal Resistance r =0 (ideal battery) real battery 9/23/15 25 7