Chapter 26. A capacitor is a device that one can use to store an electric potential energy.

Transcription

1 223 Chapter 26 A capacitor is a device that one can use to store an electric potential energy. (Spring: mechanical potential energy) Useful in electronics and microelectronics

2 (C: capacitance SI: farad) 224

3 225

4 226 The plates of a parallel-plate capacitor are separated by a distance d=1.0mm. What must be the plate area if the capacitance is to be 1.0F? Sol: Cd A = = 1.1X10 8 m 2 (large!) ε 0

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9 231 Energy density μ: The potential energy per unit volume

10 232 What is the potential energy of the two-capacitor system in problem 26-4 before and after switch S is closed = U i = C V 2 U f = C V μJ = 1 + C2V 20.0μJ (smaller!) 2 Heat (c) What is the radius R 0 of an imaginary spherical surface such that half of the stored potential energy lies within it? (a)

11 233 (b) (c) R R 0 1 du = 2 R du du du = ( u)(4πr 2 q = 8πε 0 2 dr 2 r )( dr) R R 0 dr 2 r = 1 2 R dr 2 r = R R 2R 0 R=R 0 =13.7cm κ : a numerical factor

12 Water,etc. 234

13 235

14 236

15 237 (a) (b) (c) (d) (e)

16 238 (f) Exercises:17,29,41

17 239 Chapter 27 Electric currents: Charges in motion

18 240 Current is a scalar. For historical reasons, a current arrow is drawn in the direction in which positive charge carriers would move, even if the actual charge carriers are negative and move in the opposite direction. For a constant current (SI unit: A/m 2 ) Electrons tend to drift with a drift speed νd in the direction opposite to the current.

19 (a) Uniform current density 241

20 242 (b) One end of an aluminum wire whose diameter is 2.5mm is welded to one end of a copper wire whose diameter is1.8mm. The composite wire carries a steady current I of 17ma. (a) What is the current density in each wire? (b) What is the drift speed of the conduction electrons in the copper wire? Assume that, on the average, each copper atom contributes one conduction electron. (a) J cu = 17mA d 2 π ( ) 2 =6.7X10 3 A/m 2 J J A1 cu = A A A1 cu X10 A / m = J A1 (b)

21 X10 A/ m 7 ν d = = 4.9X10 m / s = 1.8mm / h N Aρ 19 ( )(1.60X10 ) M Consider a strip of silicon has a rectangular cross section with width w=3.2mm and height h=250μm, and through which there is a uniform current I of 5.2ma. The silicon is an n-type semiconductor, having been doped with a controlled phosphorus impurity. N=1.5X10 23 m -3 (a) What is the current density in the strip? Sol: J=i/wh=6500A/m 2 (b) What is the drift speed? Sol: v d =J/ne=27cm/s R: resistance

22 ρ: resistivity 244

23 245

24 246 ( pn junction) or If an electron of mass m is placed in an electric field of magnitude E, On the average ( : the mean free time)

25 247 (a) (b)

26 248 A wire of length L=2.35m and diameter d=1.63mm carries a current I of 1.24A. The wire dissipates electrical energy at the rate P of 48.5mW. of what is the wire made? Sol: 2 4i ρl πd P= 2 ρ=2.80x10-8 Ω.M Al The valence band: the highest band that is occupied by electrons. Superconductors: Hg,... High temperature superconductivity : Paul chu,... Exercises:25,31,41

27 249 Chapter 28 A charge pump : a device that by doing work on the charge carriers maintains a potential difference between a pair of terminals. an emf device (electromotive force) Def ξ = dw dq Ideal emf device: no internal resistance Real emf device: (Battery) has internal resistance to the internal movement of charge

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30 (Kirchhoff s current law) 252

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32 (a) 254

33 255 (b) (c)

34 256 Q=CV Def. τ= RC : (the time constant ) t=τ

35 257 K=? Initial condition A capacitor of capacitance C is discharging through a resistor of resistance R (a)in terms of the time constant τ=rc, when will the charge on the capacitor be half its initial value?

36 258 (b)when will the energy stored in the capacitor half its initial value? (c)at what rate PR is thermal energy produced in the resistor during the discharging process? At what rate Pc is stored energy lost by the capacitor during the charging process (a) q = q 1 2 q 0 1 ln 2 0 e = q t RC 0 e = ln( e t RC t RC ) = t RC T=-ln(1/2)(RC)=0.69τ (b) 2 2 2t 2t q q 0 RC RC U= = e = U 0e 2C 2C 2t 1 RC U 0 = U 0e 2 ln(1/2)=-2t/rc t=0.35τ (c) t RC q PR=i 2 0 R= [ e ] RC 2 2t q 0 RC = e 2 RC d P c =du/dt= ( U 0e dt P R +P C =0 2 R 2t RC 2U 0 ) = RC e 2t RC 2q0 = RC 2 2 e 2t RC Exercises:31,43

37 259 Chapter 29 A charged plastic rod produces a vector field the electric field E in the space around it. A magnet produces a vector field the magnetic field B in the space around it. Setting up magnetic fields (1) Moving electrically charged particles, such as a current in a wire (2) Elementary particles such as electrons have an intrinsic magnetic field around them. Def. --Using the right hand rule. SI unit: Tesla (T)

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39 261 (small) a=? (K<<mc 2 ) relativistic (large)

40 We have 262

41 Hall potential difference 263

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45 267 Copper: screens the electric field. (a) relativity (b) large r In vector form:

46 268 Figure shows a length of wire with a central semicircular arc, placed in a uniform magnetic field b that points out of the plane of the figure. If the wire carries a current i, what resultant force F acts on it?

47 269 dfsinθ dl 2 θ R df 0 Sol: F 1 =F 3 =ilb df=ibdl=ib(rdθ) π F = df sinθ = ( ibrdθ )sinθ = ibr sinθdθ = 2iBR 2 0 F= F 1 + F 2 +F 3 =2iB(L+R) π 0 π 0 Electric motor

48 Right-hand rule 270

49 271 Def (magnetic dipole) ΔU=(+μB)-(- μb)=2μb Exercises:19,35,41,55