Physics Lecture 40: FRI3 DEC

Transcription

1 Physics 3 Physics 3 Lecture 4: FRI3 DEC Review of concepts for the final exam

2 Electric Fields Electric field E at some point in space is defined as the force divided by the electric charge. Force on charge at some point, by charge is given by Electric field at that point is! F! F! E k! E k R R

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4 Gauss law: Given an arbitrary closed surface, the electric flux through it is proportional to the charge enclosed by the surface. Flux! Φ E!! da ε Surface

5 A charged conductor: A conductor with a cavity:

6 Example: a charged conducting spherical sheet with a charge inside Charge of sheet : s The presence of the central charge attracts electrons to the inner surface of the metal sheet. How much charge is there on the inner surface of the sheet? Construct a spherical Gaussian surface inside the sheet. Since it is a conductor, the field there vanishes. Therefore the flux vanishes. By Gauss law, the enclosed charge should be zero. Therefore the amount of charge on the inner surface is -. Charge in the outside surface: construct a spherical Gaussian surface outside the sheet. The enclosed charge is +s. The external field will be eual to that of a point charge of value +s. Now, the external field is entirely due to the charge on the outside of the sheet (since the field due to the inner surface cancelled with that of the point charge). Therefore the amount of charge deposited on the outside is +s.

7 Definition of electric potential: Potential energy of a system per unit charge Units Units... U V V f V i U f U i f!! E ds i Units: [V] Joule Coulomb Volt [Volt] N [m] C N C V m Unit most commonly used for electric fields Alessandro Volta (745-87) ΔV ΔU ΔU ΔV evelectron-volt, the energy that an electron acuires when placed in an electric potential of V ev (.6 9 C) V.6 9 J

8 Since what matters in potential energy (and therefore in electrical potential) are differences, the potential is in general defined up to a constant. One way of fixing that constant is to declare that some point in space has zero potential. Very commonly infinity is chosen as that point. In that case we have that W V W Where is the work done by the electric field on a charged particle as it is brought from infinity to its current location. If one moves a charge across a field exerting a force on it, there are two types of work done: the one by the external force and the one by the field. Their sum will be eual to the change in the kinetic energy of the charge. If the particle is stationary before and after the move, then W app -W field ΔV.

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10 Resistivity: Resistance at a point Metal streamlines These two devices could have the same resistance R, when measured on the outgoing metal leads. However, it is obvious that inside of them go on different things. Resistivity is associated with a material, resistance with respect to a device constructed with the material. In order to uantify this, we introduce the concept of resistivity: E! ρ or, as vectors, E ρ J! J Conductivi ty : σ ρ Example: - L V A + V E, J L i A ρ Makes sense! For a given material: V i L A R A L Thicker R L ρ A Longer More resistance Less resistance

11 R L ρ A V R i and therefore : Ohm s laws V i and V ir R

12 Sketch of the solution: a,b,c) At t capacitor is discharged, so it behaves as a wire: Resistors,3 in parallel with each other and the resulting resistor in series with R. d,e,f) At tinfinity, capacitor fully charged, behaves as if the circuit is open. So there is no current in R3, and R, R are in series.

13 Magnetic versus electrostatic forces: An important difference in electric and magnetic fields is how they act on charges. For electrostatic forces :! For magnetic forces, F!! F E!! v B Charges that do not move, do not feel magnetic forces. Magnetic forces are perpendicular to both the velocity of charges and to the magnetic field (electric forces are parallel to the field). Since magnetic forces are perpendicular to the velocity, they do no work! Speed of particles moving in a magnetic field remains constant in magnitude, the direction changes. Kinetic energy is constant (no work).

14 Force between parallel wires carrying current Magnetic field due to wire where the wire is, L I I B µ I π a F Force on wire due to this field, F L I B µ LI I π a a

15 67 A length of wire is formed into a closed loop with radii a and b, and carries a current i. a) What is the value of the magnitude and direction of B at the center? b) Find the magnetic dipole moment of the circuit. Straight pieces do not contribute to B. Last class, arc: Dipole moment, µ µ iφ 4π r B µ 4 i r µ i + At center, B 4 a b π a π b πi i Area i + ( a + b a ) b

16 Electric oscillations: math E E + tot mag E elec E tot Li + C de di tot L i + dt dt C di L i + dt C ( i) d dt i d dt d Compare with: L + d x M + k x dt C dt So the math is exactly the same as in the case of mechanical oscillators if one makes the substitutions: x / C k i v L M cos( ω t + ) φ ω LC

17 Electric oscillations: graphs and energy.5.5 Time Charge Current E mag Energy in capacitor Energy in coil E ele cos( ω t + ) φ i ω sin( ω t + ) φ Li Lω sin ( ω t + φ) cos ( ω t +φ ) C C And remembering that, cos x + sin x, and ω LC. Time Etot Emag + Eele The energy is constant and eual to what we started with. C

18 Magnetic field of a solenoid No field outside, field concentrated inside. Idealized: (infinitely tightly woven, infinite) Considering Amperian loop abcd,!! b!! c!! B ds B ds + B ds + d c!! B ds + a b a!! B ds Bh Bh i inh, enc B µ i n nturns per unit length b

19 Self-inductance Suppose you have a coil with a current that changes with time. The magnetic flux in it will changewith time too. Therefore it will induce an emf in the coil! This effect is called self-induction. Changing current Changing flux 3 EMF Combining, emf L d( NΦ) dt NΦ i d( Li) dt With Faraday's law, 3 d i L d t emf N When we take a walk around a circuit to solve it, every time we find a solenoid, we add a term -L di/dt. If the current is constant, the coil is invisible (piece of wire). If we have sudden changes of current we can get large emf s with a coil. d Φ dt

20 The new term in Ampere s law (Ampere-Faraday), S S!! dφ Faraday s law told us that a E B ds µ ε + µ i changing magnetic flux produces dt an electric field. This law tells us!! B ds ( ) that a changing electric flux i + id produces a magnetic field. µ dφ Where id ε dt E, "displacement current" Example: a charging capacitor, ε AE dφ dt E i ε d dt i d de ε A ε dt dφ dt E

21 Experiments show that reflection and refraction keep the outgoing rays in the same plane as the ingoing rays and the normal of the surface and are governed by two laws: Law of reflection: the angle of incidence θ euals the angle of reflection θ. Law of refraction: n sinθ n sinθ Where n and n are called the index of refraction of media and respectively. These uantities are determined experimentally and listed in tables. For air n is very approximately. All other substances have larger indices of refraction. Snell s law. Willebrord Snell If n euals n then light travels straight. If n is smaller than n then the refracted angle is smaller than the incident, otherwise larger. It can be so large that the light is actually reflected. It can never be so large that it will go beyond the normal. René Descartes

22 Electromagnetic waves are able to transport energy from transmitter to receiver (example: from the Sun to our skin). The amount of power transported by the wave and its direction is uantified by a vector called Poynting vector.! S µ! E! B S EB E µ cµ The units are power per unit area, i.e. Watt/m John Henry Poynting (85-94) For a wave since E is perpendicular to B In a wave, the fields change with time in a fixed way. Therefore the Poynting vector changes too. A better measure of the amount of energy is obtained by averaginng the Poynting vector over one wave cycle. The resulting uantity is called intensity

23 I S E Em sin ( kx ωt) cµ cµ I cµ E m The average of sin over one cycle is /. Engineers commonly use the term root mean suare value of a uantity, Em Erms Em. 44Em I E rms c µ Both fields have the same energy density B E εe ε( cb) ε u B εµ u

24 The intensity of a wave is power per unit area. If one has a source that emits isotropically (eually in all directions) the power emitted by the source pierces a larger and larger sphere as the wave travels outwards. Therefore, I P s 4πr So the power per unit area decreases as the inverse of distance suared. Waves not only carry energy but also momentum. The effect is very small (we don t ordinarily feel pressure from light). If light is completely absorbed during an interval Δt, the momentum transferred is given by, Δu Δp And if light is reflected, one gets double this amount. c

25 When polarized light hits a polarizing sheet, only the component of the field aligned with the sheet will get through. E y E cos( θ ) And therefore: I I cos θ Polarized sunglasses operate on this formula. They cut the horizontally polarized light from glare (reflections on roads, cars, etc).

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