Physics 9 Monday, March 19, 2012

Transcription

1 Physics 9 Monday, March 19, 2012 learningcatalytics.com class session ID: Electricity: lots of new words & ideas, so ask lots of questions! Today: Gauss s law; electric potential (volts!); capacitance. Soon: electric current (amps!) If you find the past two Giancoli chapters confusing or unsatisfying, you can try (optional) 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 your comments on 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 ).

2 Motion in electric field. Remember that the vector sum of forces acting ON an object causes the object to accelerate: m a = F In an electric field E, the force F on an object with charge q is F = q E If the force F = q E is not balanced by any other force, a charged object will accelerate in an electric field: a = q m E If some other force F other is also acting, then (Questions 1, 2.) a = q m E + 1 m F other

3 Electric field hockey!! phet.colorado.edu/en/simulation/electric-hockey

4 E.F.H can draw the electric field e.g. from HW problem 2. (The black + is the test charge and doesn t contribute to E.) (Question 3.)

5 Flux of electric field lines. E( R) = 1 4πɛ 0 dx dy dz ρ( r) r R r R 3 The way to avoid using this cumbersome equation is: First to remember that E is proportional to the density of electric field lines. More closely spaced lines bigger E. Then to think of the field lines as something flowing (or radiating, like the intensity of light or sound) out from the (+) charges and into the ( ) charges. Then to use the amazing trick that the total flux of E through a hypothetical closed surface is proportional to the total charge enclosed by that surface. Just as we earlier added up the intensity over an enclosing surface area to get total power, we will now add up the electric flux over an enclosing surface to get Q enc /ɛ 0.

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7 Gauss s Law. The flux of electric field E (uppercase phi for flux ): Φ E = Q enclosed ɛ 0 You draw a hypothetical closed surface (closed = air tight, no missing sides). Then you add up over the whole surface area the normal ( ) component of E. E da = Q enclosed ɛ 0 surface To make math easy, choose your shape so that every side is either to E (counts 100%) or to E (counts 0). Since E = 0 inside conductor (for static charges), you often choose one surface to be inside a conductor, if there is one. Examples (on board): E outside point charge; E outside charged sphere; E between parallel plates.

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9 E at distance R from a single point charge

10 ~ at distance r from conducting sphere of radius R E

11 Demonstrate Faraday cage concept using grass seeds. Notice that E = 0 inside the metal cage, because Q enclosed = 0. (Question 4.)

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14 E da = Q enclosed surface ɛ 0 Point charge: E(r) Q = 4πr 2 ɛ 0 Conducting sphere (all charge stays on outer surface, at r = R): E(r) = Q 4πr 2 ɛ 0 (r > R) E(r) = 0 (r < R) Uniform line charge (infinite length): E(r) = (Q/length) 2πrɛ 0 Uniform plane of charge (infinite area): E = (Q/area) 2ɛ 0

15 Parallel conducting plates, +Q and Q

16 Work and potential energy in electrostatics. If I move a charged particle along the direction of electric field E, I decrease the particle s potential energy: U = q E d r We define electric potential (a.k.a. voltage) as the potential energy per unit charge: V = U = E d r q So E is the derivative (technically the gradient ) of V : E x = dv dx, E y = dv dy, E z = dv dz So newtons coulomb equals volts meter. (Equivalent units for E.) Topo map: V is like elevation and E points downhill.

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20 Back to 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 > R : r V (r) = r R : V (r) = Q 4πɛ 0 R notice that V Q Q dr 4πɛ 0 r 2 = Q 4πɛ 0 [ 1 ] r = Q r 4πɛ 0 r

21 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 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 = C V.