Physics 9 Friday, March 7, 2014

Transcription

1 Physics 9 Friday, March 7, 2014 HW7 due now. I ll post HW8 online this weekend, due Friday after break (3/21): will mostly cover voltage. Read all of Ch26 (charge separation & storage) by Monday after break (3/17). Remember to answer both sets of reading questions: concepts for 3/5 (spring break OK), eqns for 3/17. Ch26 is about devices that move positive & negative charge away from one another (e.g. a battery) or store them in that separated state (e.g. a capacitor). These will be important later for pushing electrons around in electric circuits. Ch22: charges, opposites attract, likes repel, F = kq 1 q 2 /r 2 12 Ch23: E field = force per unit charge; superposition Ch24: E field lines, flux, enclosed charge, Gauss s Law Ch25: voltage = electric potential energy per unit charge Ch26: van de Graff generator, batteries, capacitors Ch27 (for March 19) is about magnets!

2 Let s review Wednesday s quiz first Question: what is a heat engine? (A) It s a device that uses mechanical (or electrical) energy (called work ) to move heat from a cooler place to a warmer place (which is the opposite of the direction heat naturally moves). Examples include refrigerators and air conditioners. (B) It s a device that takes in heat from a hotter place, converts a fraction of that heat into useful mechanical energy ( work ), and exhausts wasted heat to a cooler place. Examples include a car s internal combustion engine, an airplane s jet engine, a diesel engine, a fuel-buring power plant (where electricity is generated), and an old-fashioned steam engine. (C) Both of the above are called heat engines. (D) None of the above.

3 Wednesday s quiz was difficult because (A) We didn t cover steady devices very thoroughly. (B) The steady device ideas are just more difficult/abstract. (C) This was a really busy week (exams, papers, critiques), so I didn t have as much time to review for the quiz as usual. (D) I filled my brain with brand new electrical ideas, so some of the heat stuff got pushed aside to make room. (E) When I read heat engine, my brain was thinking heat pump instead. The word engine would have been more clear than the phrase heat engine. (F) Physics is getting boring. Is it time for spring break yet? (G) Some other reason?

4 The efficiency of an engine (what you want it to do, divided by what you have to pay for) equals (A) Q input W (B) Q input Q output (C) Qoutput Q input (D) Qoutput W (E) (F) W Q input W Q output

5 An engine works most efficiently when (A) T input = T output (B) T input T output (C) T input T output (D) T input = 273 C

6 If you use a heat pump to heat your house, the C.O.P. is defined as (A) (B) (C) (D) (E) Q input W Q output Q input Q output W W Q input W Q output

7 A heat pump moves heat from a cooler place to a warmer place. Its performance is best (i.e. coefficient of performance is largest) when the temperature difference between these two places is (A) as large as possible (B) not very large (C) measured on the Kelvin scale

8 I want to heat my house. If I take one kilojoule of electrical energy and use it to run a heat pump whose C.O.P. is 10, the amount of thermal energy the pump moves into my house as a result is (A) 0.1 kj (B) 1 kj (C) 10 kj (D) kj

9 put positive sheet up above (and parallel to) negative sheet? Where do the E field lines add up, and where do they cancel each other out?

10 The most commonly used way to create a uniform electric field is to use the area between two large, parallel, oppositely-charged planes of uniform charge-per-unit-area.

11 Notice that if you do this, a positive particle will fall in the direction that E points, just as a rock will fall in the direction gravity points toward Earth s surface. To lift up a positive particle, you would have to add energy (do + work).

12 Electrostatic potential ( voltage ) is analogous to altitude. Gravity points in the direction in which altitude decreases most quickly. E points in the direction in which voltage decreases most quickly. Equipotential lines are perpendicular to E.

13 The potential difference between point a and point b is minus the work-per-unit-charge done by the electric field in moving a test particle from a to b. V ab = 1 b b F E d q l = E d l a a More intuitively, V ab is (plus) the work-per-unit-charge that an external agent (like me) would have to do to move a particle from a to b. I would be working against the electric field to do this. But a much easier-to-remember definition of voltage is electric potential energy per unit charge. Just as E is electric force per unit charge, V is electric potential energy per unit charge. V = UE q Moving a positive particle to higher V means moving it to a position of higher electric potential energy.

14 Near Earth s surface, gravitational potential energy is U G = m g h G.P.E. per unit mass would be just (U/m) = gh, which is proportional to altitude. Moving an object (no matter what mass) along a contour of equal gh does not require doing any work against gravity, and does not change the object s G.P.E. In a uniform downward-pointing electric field, electric potential energy is U E = q E y E.P.E. per unit charge would be just V = (U/q) = E y. So if E is uniform and points down, then potential (or voltage ) V is analogous to altitude. Moving perpendicular to E does not require doing any work against E, and does not change E.P.E. So equipotential lines (constant V) are always perpendicular to E.

15 I am standing in a uniform electric field, of magnitude 1 N/C, which points downward. I climb up 1 meter. What is the potential difference, V 1 2 = V 2 V 1, between my old location and my new location? (Note: 1 N/C is the same as 1 volt per meter.) (A) V 1 2 = +1 volt (B) V 1 2 = 1 volt (C) V 1 2 = 0 volts

16 2 V 12 = E d l = E y dy = E y (y 2 y 1 ) V 12 = (E y ) (y 2 y 1 ) = ( 1 N/C) (+1 m) = +1 volt It makes sense that moving from 1 to 2 increases V, because the electric potential energy per unit charge is larger at the final position. The work-per-unit-charge done by the external agent (me) is positive. (That s the opposite of the work-per-unit-charge done by the electric field.)

17 The van-de-graff generator accumulates surplus electrons on its metal sphere. A rubber belt rubs against a piece of wool at the lower base, carries electrons up to the metal sphere, and deposits them there. Figure 26.3 has a nice diagram of a vdg, but ours actually accumulates negative charge, not positive charge. To agree with the book s figure, let s pretend that our vdg also accumulates positive charge. What would the electric field lines look like around the positively charged metal sphere? How much surplus charge can accumulate on the sphere (R = 0.12 m) before the electric field near the sphere s surface reaches the breakdown field for dry air, E breakdown = V/m?

18 Electric field lines: (If we pretend that positive charge collects on the sphere. In our own vdg it will turn out to be negative.) By the way, what would the equipotential surfaces look like on this diagram? What does E look like within the thin metallic part of the metal spherical shell? What does E look like in the empty region enclosed by the metal shell? Using Gauss s Law, you found last night (for r > R) that 4πr 2 E = Q E = Q ɛ 0 4πɛ 0 r 2 = kq r 2 So just outside the sphere, at r 0.12 m, we get E = V/m if Q = 4.8 µc ( protons).

19 Just outside the sphere, at r 0.12 m, we get E = V/m if Q = 4.8 µc. So if the sphere is fully charged up (at the threshold for making a spark), then for r > R = 0.12 m we have E(r) = kq r 2 How do we figure out the potential difference ( voltage ) between the Earth (ground) and the surface of the sphere? How many volts? Will it turn out to be a frighteningly large voltage? Let s move a positive test charge in from infinity (i.e. very far away) to the surface: V R = = kq R R E d l = + 1 dr = kq r 2 [ 1 R ] r R ˆr E d l = + = kq R [( 1 E r dr = + ) = V = 0.36 MV ( 1 R R kq r 2 )] = kq R That s like a quarter of a million AAA flashlight batteries! But it s very little charge, so the total stored energy is small (about 2 J). dr

20 Take another look at this description when you finish reading Chapter 26, and think about the electric field inside and outside of the sphere. How can we see that our own vdg actually charges up with surplus electrons (negative charge)? What do its field lines look like? Is it at higher or lower potential ( voltage ) than the Earth?

21 Physics 9 Friday, March 7, 2014 HW7 due now. I ll post HW8 online this weekend, due Friday after break (3/21): will mostly cover voltage. Read all of Ch26 (charge separation & storage) by Monday after break (3/17). Remember to answer both sets of reading questions: concepts for 3/5 (spring break OK), eqns for 3/17. Ch26 is about devices that move positive & negative charge away from one another (e.g. a battery) or store them in that separated state (e.g. a capacitor). These will be important later for pushing electrons around in electric circuits. Ch22: charges, opposites attract, likes repel, F = kq 1 q 2 /r 2 12 Ch23: E field = force per unit charge; superposition Ch24: E field lines, flux, enclosed charge, Gauss s Law Ch25: voltage = electric potential energy per unit charge Ch26 (for 3/17): van de Graff generator, batteries, capacitors Ch27 (for 3/19) is about magnets! Have a safe and happy spring break!