Physics 9 Wednesday, March 21, 2012

Transcription

1 Physics 9 Wednesday, March 21, 2012 learningcatalytics.com class session ID: today: potential (voltage) capacitance soon: electric current circuits HW10 covers only chapters G16 and G17 ( E, V, and C) If you find last week s Giancoli chapters unsatisfying, you can try reading the conceptual parts of Eric s chapter 23 (E field), 24 (Gauss s law), 25 (potential), and 26 (capacitance). I put them on the usual nb.mit.edu page. If you do read any of these chapters and send me thoughtful answers to What did you find interesting / confusing / helpful / etc., you can earn extra credit comparable to one night s reading credit per chapter. So reading one extra chapter could raise your final grade e.g. from 95.0% to 95.3%. Likewise for extra Muller chapter ( heat & atoms ). A few people s earning extra credit will not hurt anyone else s grade. There is no curve. So if you don t feel like reading extra chapters, there is no pressure to do so.

2 Good answer for today s reading The electric current is analogous to the amount of water flowing through the stream subject to the force of gravity. The greater the height difference between the top and bottom of the hill, the faster the flow of water through the stream; and likewise the greater the potential difference in a circuit, the greater the current. Electric potential is analogous to the potential energy of an object acted on by gravity - potential energy of the water in the stream increases the greater the height, or the farther up the hill it is, and a greater difference in electric potential creates a greater electric current flow. I = V/R = 1.5 volt / 1500 ohm = 0.001A. This will be a Direct Current, not an AC. (But in class today, we re still working on last Friday s reading!)

3 Imagine the shape of a hill. If you let go of a ball, it starts rolling downhill. The net force points in the direction in which potential energy decreases most rapidly. If x points east, y points north, and z points up, then U(x, y) = mg z(x, y) and F x = du dx F y = du dy Bigger slope bigger force. (Question 1.)

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6 (Question 2.)

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8 In electricity, potential (a.k.a. voltage) is like the elevation on your topo map. Electric potential is potential energy per unit charge for a small test charge q V = U q The electric field always points in the downhill direction: the direction in which V decreases most rapidly. E x = dv dx, E y = dv dy, E z = dv dz So I can measure the strength of an electric field equivalently as. (These two units are equal.) newtons coulomb or volts meter Electric field E measures force per unit charge how rapidly potential varies with position (e.g. HW10 # 6) direction in which potential decreases most rapidly

9 (Question 3.)

10 Computing V near a single charge Remember that the electric field at a distance r from a single point charge q is E = q 4πɛ 0 r 2 ˆr Since E is the derivative of V, we can get V by integrating E.

11 Computing V near a single charge Remember that the electric field at a distance r from a single point charge q is E = q 4πɛ 0 r 2 ˆr Since E is the derivative of V, we can get V by integrating E. If we take V = 0 at r =, then the potential at a distance r from a point charge q is r r q V (r) = dr E(r) = dr 4πɛ 0 r 2 = q [ ] 1 r 4πɛ 0 r V (r) = (For example, HW10, problem 1.) q 4πɛ 0 r

12 Computing V near several charges If the point (x, y, z) is a distance d i away from each of N point charges q i, then the individual contributions just add up: V (x, y, z) = N i=1 q i 4πɛ 0 d i where d i = (x x i ) 2 + (y y i ) 2 + (z z i ) 2 So you can see that computing V ( r) for a set of charges is much less tedious than computing E( r) for a set of charges. Once you know V (x, y, z), you can use it to figure out E(x, y, z). (For example, HW10, problem 5.)

13 Charged conducting sphere (of radius R) r > R : 4πr 2 E = Q enc /ɛ 0 = Q/ɛ 0 r < R : 4πr 2 E = Q enc /ɛ 0 = 0 So E(r) = 0 for r < R, and E(r) = Q 4πɛ 0 r 2 for r > R. E x = dv dx qe x = q dv dx F x = du dx Let V ( ) = 0. Then r Q r > R : V (r) = dr 4πɛ 0 r 2 = Q 4πɛ 0 r R : V (r) = Q 4πɛ 0 R [ 1 ] r = Q r 4πɛ 0 r Notice that V Q. Capacitance C = Q/V = 4πɛ 0 R (Question 4. Then van de Graff generator.)

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15 Capacitance capacity to store electric charge

16 Parallel conducting plates, +Q and Q

17 Two large, parallel conducting plates, of area A Put charge Q at x = 0 and charge +Q at x = d. E = Q/area ɛ 0 = constant between plates E x = Q/area ɛ 0 = Q ɛ 0 A Let V = 0 at x = 0. x ( V (x) = E x dx = Q ) [ ] x x = Qx 0 ɛ 0 A 0 ɛ 0 A V (d) = Qd ɛ 0 A So V between the two plates is V = Qd ɛ 0 A. Notice that V Q. Capacitance C = Q V = ɛ 0A d