Chapter 8 Electrical Energy and Capacitance Work •Recall: A force is conservative if the work done on an object when moving from A to B does not depend on the path followed. Consequently, work was defined as: W=PEi-PEf=-PE • Work done BY gravity decreases PE. •This was derived for a gravitational force, but as we saw in the previous chapter, gravitational and Coulomb forces are very similar: Fg=Gm1m2/r122 with G=6.67x10-11 Nm2/kg2 Fe=keq1q2/r122 with ke=8.99x109 Nm2/C2 Hence: The Coulomb force is a conservative force. Work & Potential Energy •consider a charge +q moving in an E field from A to B over a distance d. We can ignore gravity (why?). •What is the work done by the field? •What is the change in PE? •If initially at rest, what is its speed at B? •WAB=Fdcos ; = 0, so •WAB=Fd •WAB=qEd (since F=qE) •work done BY the field ON the charge (W is positive) • PE=-WAB=-qEd : negative, so the potential energy has decreased. •Conservation of energy: •PE+KE=0 •KE=1/2m(vf2-vi2) •1/2mvf2=qEd •v=(2qEd/m) work & potential energy II •WAB=-qEd ; negative, so work •Consider the same situation for a charge of –q. •Can it move from A to B without an external force being applied, assuming the charge is initially (A) and finally (B) at rest? must be done by the charge. This can only happen if an external force is applied. •Note: if the charge had an initial velocity the energy could come from the kinetic energy (I.e. it would slow down) •If the charge is at rest at A and B: external work done: qEd •If the charge has final velocity v then external work done: W=1/2mvf2+|q|Ed Conclusion •In the absence of external forces, a positive charge placed in an electric field will move along the field lines (from + to -) to reduce the potential energy •In the absence of external forces, a negative charge placed in an electric field will move along the field lines (from - to +) to reduce the potential energy +++++++++++++ -------------------- question P Q •a positive charge initially at rest at P moves to Q. Will it follow the shortest route (as indicated by the dashed arrow) a) yes b) no •Answer: b) since the initial force is directed along the field line •Will the change in potential energy of the charge at Q be different depending on which path is taken from P to Q? a) yes b) no c) depends on whether the velocity of the charge at Q is different depending on the path d) depends on the external forces applied •Answer: b) no The potential energy depends only on the location in the Efield, not on the path taken. question 2m Y 1m ---------------------- X •a negatively charged (-1 C) mass of 1 g is shot diagonally in an electric field created by a negatively charge plate (E=100 N/C). It starts at 2 m distance from the plate and stops 1 m from the plate, before turning back. What was the initial velocity in the direction along the field lines? answer •Note: the direction along the surface of the plate does not play a role(there is no force in that direction!) Kinetic energy balance •Initial kinetic energy: 1/2mv2=0.5*0.001*(vx2+vy2) •Final (at turning point) kinetic energy: 0.5*0.001*vx2 •Change in kinetic energy: KE=-0.5*0.001*vy2 =-5x10-4vy2 Potential energy balance •Change in Potential energy: PE=-qEd=1x10-6*100*1=+10-4 J •Conservation of energy: PE+KE=0 •vy=0.44 m/s so -5x10-4vy2+10-4=0 Electrical potential •The change in electrical potential energy of a particle of charge Q in a field with strength E over a distance d depends on the charge of the particle: PE=-QEd •For convenience, it is useful to define the difference in electrical potential between two points (V), that is independent of the charge that is moving: V= PE/Q=-|E|d •The electrical potential difference has units [J/C] which is usually referred to as Volt ([V]). It is a scalar •Since V= -Ed, so E= -V/d the units of E ([N/C] before) can also be given as [V/m]. They are equivalent, but [V/m] is more often used. Electric potential due to a single charge V + V=keq/r r 1C •the potential at a distance r away from a charge +q is the work done in bringing a charge of 1 C from infinity (V=0) to the point r: V=keq/r •If the charge that is creating the potential is negative (-q) then V=-keq/r •If the field is created by more than one charge, then the superposition principle can be used to calculate the potential at any point example 1 1 m 2 -2 C +1C r a) what is the electric field at a distance r? b) what is the electric potential at a distance r? - a) E=E1 E2=ke(Q1/r2)-ke(Q2/[1-r]2)=ke(1/r2)-ke(-2/[1-r]2)= ke(1/r2+2/[1-r]2) Note the -: E is a vector b) V=V1+V2=ke(Q1/r)+ke(Q2/[1-r])=ke(1/r-2/[1-r]) Note the +: V is a scalar question • a) b) c) d) a proton is moving in the direction of the electric field. During this process, the potential energy …… and its electric potential …… increases, decreases decreases, increases increases, increases decreases, decreases + + - PE=-WAB=-qEd, so the potential energy decreases (proton is positive) V= PE/q, so the electric potential that the proton feels decreases Note: if the proton were exchanged for an electron moving in the same direction, the potential energy would increase (electron is negative), but the electric potential would still decrease since the latter is independent of the particle that is moving in the field example •a particle (q1) with a charge of +4.5C is fixed in space. From a distance of 3.70 cm, a particle (q2) of mass 6.9 g and charge –3.10 C is fired with initial velocity of 60 m/s towards to fixed charge. What is its velocity when it is 1 cm away from q1? Use conservation of energy: KE+PE=0 At the start: PE=q2V=q2 x (keq1/r)=-3.39 J (r=0.037 cm) At collision: PE=q2V=q2 x (keq1/r)=-12.55 J (r=0.01 cm) The potential energy has decreased (more negative), so the kinetic energy has increased by (-3.39-(12.55))=9.16 J KE=0.5mv2 so, KEfinal-KEinitial=0.5x0.0069vfinal2-0.5x0.0069x602=9.16 J solve for vfinal = 79.1 m/s question +1 A -2 +1 -1 +1 B +1 -1 1) the electric potential at A is a) zero b) non-zero 2) the electric field at B is a) zero b) non-zero 3) a + particle at A would a) move b) not move 1) the electric potential at B is a) zero b) non-zero 2) the electric field at B is a) zero b) non-zero 3) a + particle at A would a) move b) not move equipotential surfaces compare with a map A capacitor +++++++++++++ +Q d symbol for capacitor when used in electric circuit -Q ------------------- •is a device to create a constant electric field. The potential difference V=Ed •is a device to store charge (+ and -) in electrical circuits. •the charge stored Q is proportional to the potential difference V: Q=CV •C is the capacitance, units C/V or Farad (F) •very often C is given in terms of F (10-6F), nF (10-9F) pF (10-12F) •Other shapes exist, but for a parallel plate capacitor: C=0A/d where 0=8.85x10-12 F/m and A the area of the plates electric circuits: batteries •The battery does work (e.g. using chemical energy) to move positive charge from the – terminal to the + terminal. Chemical energy is transformed into electrical potential energy. •Once at the + terminal, the charge can move through an external circuit to do work transforming electrical potential energy into other forms Symbol used in electric circuits: + - Our first circuit 10nF 12V •The battery will transport charge from one plate to the other until the voltage produced by the charge build-up is equal to the battery charge •example: a 12V battery is connected to a capacitor of 10 nF. How much charge is stored? •answer Q=CV=10x10-9 x 12V=120 nC •if the battery is replaced by a 300 V battery, and the capacitor is 2000F, how much charge is stored? •answer Q=CV=2000x10-6 x 300V=0.6C •We will see later that this corresponds to 0.5CV2=90 J of energy, which is the same as a 1 kg ball moving at a velocity of 13.4 m/s capacitors in parallel C1=10nF A C2=10nF 12V B At the points the potential is fixed to one value, say 12V at A and 0 V at B This means that if the capacitances C1 and C2 are equal they must have the same charge stored and the total charge stored is Q=Q1+Q2. •We can replace C1 and C2 with one equivalent capacitor: Q1=C1V & Q2=C2V is replaced by: Q=CeqV since Q=Q1+Q2 , C1V+C2V=CeqV so: •Ceq=C1+C2 •This holds for any combination of parallel placed capacitances Ceq=C1+C2+C3+… •The equivalent capacitance is larger than each of the components capacitors in series A B C1=10nF 12V C2=10nF The voltage drop of 12V is over both capacitors. V=V1+V2 The two plates enclosed in are not connected to the battery and must be neutral on average. Therefore the charge stored in C1 and C2 are the same •we can again replace C1 and C2 with one equivalent capacitor but now we start from: V=V1+V2 so, V=Q/C1+Q/C2=Q/Ceq and thus: 1/Ceq=1/C1 + 1/C2 •This holds for any combination of in series placed capacitances 1/Ceq=1/C1+1/C2+1/C3+… •The equivalent capacitor is smaller than each of the components question • Given three capacitors of 1 nF, an capacitor can be constructed that has minimally a capacitance of: a) 1/3 nF b) 1 nF c) 1.5 nF d) 3 nF The smallest possible is by putting the three in series: 1/Ceq=1/C1+1/C2+1/C3=1+1+1=3 so Ceq=1/3 nF a more general case: what is the equivalent C C4 C3 C5 C6 STRATEGY: replace subgroups of capacitors, starting at the smallest level and slowly building up. C2 C1 12V •step 1: C4 and C5 and C6 are in parallel. They can be replaced by once equivalent C456=C4+C5+C6 step II C3 C456 C1 C2 12V •C3 and C456 are in series. Replace with equivalent C: 1/C3456=1/C3+1/C456 so C3456=C3C456/(C3+C456) •C1 and C2 are in series. Replace with equivalent C: 1/C12=1/C1+1/C2 so C12=C1C2/(C1+C2) step III C3456 C12 12V C123456 12V •C12 and C3456 are in parallel, replace by equivalent C of C123456=C12+C3456 problem C3 A C4 C5 C2 C1 12V B C1=10nF C2=20nF C3=10nF C4=10nF What is Vab? C5=20nF •V12=V345=12V •C45=C4+C5=10nF+20nF=30nF •C345=C3C45/(C3+C45)=300/40=7.5nF •Q345=V345C345=12V*7.5nF=90nC •Q45=Q345 •V45=Q45/C45=90nC/30nF=3V •check V3=Q3/C3=Q345/C3=90nC/10nF=9V V3+V45=12V okay! energy stored in a capacitor +++++++++++++ +Q V V -Q ------------------- Q Q •the work done transferring a small amount Q from – to + takes an amount of work equal to W=VQ •At the same time, V is increased, since V=(Q+Q/C) •The total work done when moving charge Q starting at V=0 equals: W=1/2QV=1/2(CV)V=1/2CV2 •Therefore, the amount of energy stored in a capacitor equals: EC=1/2CV2 example •A parallel-plate capacitor is constructed with plate area of 0.40 m2 and a plate separation of 0.1mm. How much energy is stored when it is charged to a potential difference of 12V? answer: First calculate C=0A/d=8.85x10-12 x 0.40 / 0.0001=3.54x1 Energy stored: E=1/2CV2=0.5x3.54x10-8x122=2.55x10-6 J Now let’s assume a 2000F capacitor being charged with a 300V battery: E=1/2CV2=90J This is similar to a ball of 1 kg being fired at 13.4 m/s!! capacitors II +++++++++++++ +Q A d -Q ------------------- material vacuum 1.00000 air 1.00059 glass 5.6 paper 3.7 water 80 •the charge density of one of the plates is defined as: =Q/A •The equation C=0A/d assumes the area between the plates is in vacuum (free space) •If the space is replaced by an insulating material, the constant 0 must be replaced by 0 where (kappa) is the dielectric constant for that material, relative to vacuum •Therefore: C=0A/d why does inserting a plate matter? • molecules, such as water, can/is be polarized • when placed in an E-field, the orient themselves along the field lines; the negative plates attracts the positive side of the molecules •near to positive plate, net negative charge is collected; near the negative plate, net positive charge is collected. •If no battery is connected, the initial potential difference V between the plates will drop to V/. •If a battery was connected, more charge can be added, increasing the capacitance from C to C problem •An amount of 10 J is stored in a parallel plate capacitor with C=10nF. Then the plates are disconnected from the battery and a plate of material is inserted between the plates. A voltage drop of 1000 V is recorded. What is the dielectric constant of the material? answer: step 1: Ec=1/2CV2 so 10=0.5x 10x10-9 V2, V=44721 V step 2: after disconnecting and inserting the plate, the voltage over the capacitor is equal to Voriginal/ So: (44721-1000)=44721/ =1.023 problem • a) b) c) d) An ideal parallel plate capacitor is connected to a battery and becomes fully charged. The capacitor is then disconnected and the separation between the plates is increased in such a way that no charge leaks off. The energy stored in the capacitor has increased decreased not changed become zero answer: Ec=1/2CV2 with C=0A/d. If d increases, C becomes smaller. The charge remains the same and Q=CV. So, if C becomes smaller, V becomes larger by the same factor. Rewrite: Ec=1/2CV2=1/2QV. Since Q is constant and V goes up, Ec must increase. Remember •Electric force acting on object 1 (or 2): F=keq1q2/r122 •Electric field due to object 1 at a distance r: E=keq1/r2 •Electric potential at a distance r away from a charge q1: V=keq1/r Producing a Sound Wave •Sound waves are longitudinal waves traveling through a medium. •A tuning fork can be used as an example of producing a sound wave. •A tuning fork will produce a pure musical note (i.e. one with a Sinusoidal Waveform). Using a Tuning Fork to Produce a Sound Wave •As the tine swings to the right, it forces the air molecules near it closer together •This produces a high density area in the air. •This is an area of compression Using a Tuning Fork, cont. •As the tine moves toward the left, the air molecules to the right of the tine spread out •This produces an area of low density •This area is called a rarefaction Using a Tuning Fork, final •As the tuning fork continues to vibrate, a succession of compressions and rarefactions spread out from the fork •A sinusoidal curve can be used to represent the longitudinal wave •Crests correspond to compressions and troughs to rarefactions Tuning Fork (DEMO) Categories of Sound Waves The physical quantity that determines the pitch of sound is the frequency. •Audible waves •Normal range of hearing of the human ear. •Normally between 20 Hz to 20,000 Hz •Infrasonic waves •Frequencies are below the audible range. •Eexample: Earthquakes •Ultrasonic waves •Frequencies are above the audible range. •Example: Dog whistle. Applications of Ultrasound •Widely used as a diagnostic and treatment tool in medicine. •Ultrasonic Sound Imaging: •Image small body organs/organs that can’t be resolved by X-rays (e.g Liver, spleen). •Observe foetuses in the womb. •Ultrasonic flow meter to measure blood flow (Doppler Effect). •Cavitron Ultrasonic Surgical Aspirator (CUSA) used to surgically remove brain tumors. Speed of Sound v B v Y vair r r Liquids and Gases: B is bulk modulus, r is density (mass/volume) Solids: Y is Young’s modulus m T (331 ) s 273 K • • 331 m/s is v at 0° C; T is the absolute temperature. Speed of sound is higher in solids than in fluids (coz of atomic/molecular spacing). Example 6.1 John Brown hits a steel railroad rail with a hammer. Betsy Brown, standing one mile (1609 m) down the track, hears the bang through the cool 32 F air while her twin sister Boopsie is lying next to her and hears the bang through the steel by placing her ear on the track. DATA: Ysteel=2.0x1011 Pa, rsteel=7850 kg/m3 What is the time difference between the moments when Betsy and Boopsie hear the bang? 4.54 s Intensity of Sound Waves E P I A t A Power Area SI units are W/m2 Intensity is proportional to square of amplitude (pressure modulation) Various Intensities of Sound The physical quantity that determines the loudness of sound is Intensity. •Threshold of hearing •Faintest sound most humans can hear •About 1 x 10-12 W/m2 •Threshold of pain •Loudest sound most humans can tolerate •About 1 W/m2 •The ear is a very sensitive detector of sound waves •It can detect pressure fluctuations as small as about 3 parts in 1010 Decibel Scale Sensation is logarithmic 10 log10 I Io I I 0 10 /10 • I0 is threshold of hearing (0 dB) • Threshold of Pain is therefore 120 dB Intensity vs. Intensity Level • INTENSITY is P/A, W/m2 • INTENSITY LEVEL is in decibels (dimensionless) Example 6.2 A noisy machine in a factory produces a sound with a level of 80 dB. How many machines can the factory house without exceeding the 100-dB limit? 100 machines Spherical Waves • • A spherical wave propagates radially outward from the oscillating sphere. Energy propagates equally in all directions. P I 4 r 2 2 I r 1 22 I 2 r1 Exercise A train sounds its horn as it approaches an intersection. The horn can just be heard at a level of 50 dB by an observer 10 km away. Treating the horn as a point source and neglect any absorption of sound by the air or ground, a) What is the average power generated by the horn? a) 126 W b) What intensity level of the horn’s sound is observed by someone waiting at an intersection 50 m from the train? b) 96 dB Doppler Effect • A change in the frequency is experienced by an observer due to relative motion between a source of waves and an observer. •When the source and the observer are moving toward each other, the observer hears a higher frequency. •When the source and the observer are moving away from each other, the observer hears a lower frequency. Doppler Effect, Moving Observer (stationary source) When not moving, f v When observer moving, f ' (v vobs ) v vo ƒ' ƒ v Fig 14.8, p. 435 Slide 12 If observer moves away from source: v vo ƒ' ƒ v Fig 14.9, p. 436 Slide 13 Example 14.5 Mary is riding a roller coaster. Her mother who is standing on the ground behind her yells out to her at a frequency of 1000 Hz, but it sounds like 920 Hz. (v=343 m/s) What is Mary’s speed? 27.4 m/s Doppler Effect, Case 2 (Source in Motion) •As the source moves toward the observer (A), the wavelength appears shorter and the frequency increases. •As the source moves away from the observer (B), the wavelength appears longer and the frequency appears to be lower Doppler Effect Source in Motion ' vs T vs v 1 vs v f ' v' v f' f v vs ' Doppler Effect, Source in Motion Approaching source: v f' f v vs Source leaving: v f' f v vs Example 6.4 An train has a brass band playing a song on a flatcar. As the train approaches the station at 21.4 m/s, a person on the platform hears a trumpet play a note at 3520 Hz. DATA: vsound = 343 m/s a) What is the true frequency of the trumpet? a) 3300 Hz b) What is the wavelength of the sound? b) 9.74 cm c) If the trumpet plays the same note after passing the platform, what frequency would the person on the platform hear? c) 3106 Hz Application of Doppler Effect: Speed Radar v f' f v vs •By measuring the frequency shift in the incident and reflected wave, the velocity of the moving object can be known. Shock Waves •A shock wave results when the source velocity exceeds the speed of the wave itself. •The circles represent the wave fronts emitted by the source Shock Waves, final •Shock waves carry energy concentrated on the surface of the cone, with correspondingly great pressure variations •A jet produces a shock wave seen as a fog Interference of Sound Waves •Sound waves interfere •Constructive interference occurs when the path difference between two waves’ motion is zero or some integer multiple of wavelengths •path difference = nλ •Destructive interference occurs when the path difference between two waves’ motion is an odd half wavelength •path difference = (n + ½)λ Interference of Sound Waves Assume sources “a” and “b” are “coherent” (producing waves of the same wavelength). If observer is located ra and rb from the two sources, Source a ra Source b rb ra rb n for maximum ra rb (n 1 2) for minimum Observer Exercise 14.12 A pair of speakers separated by 1.75 m are driven by the same oscillator at a frequency of 686 Hz. An observer starts at one of the speakers and walks on a path that is perpendicular to the separation of the two speakers. (Assume vsound = 343 m/s) a) What is the position of the last intensity maximum? a) 2.81 m b) What is the position of the last intensity minimum? b) 6.00 m c) What is the position of the first intensity maximum? c) 27 cm Standing Waves •When a traveling wave reflects back on itself, it creates traveling waves in both directions •The wave and its reflection interfere according to the superposition principle •With exactly the right frequency, the wave will appear to stand still •This is called a standing wave Standing Waves, cont •A node occurs where the two traveling waves have the same magnitude of displacement, but the displacements are in opposite directions •Net displacement is zero at that point •The distance between two nodes is ½λ •An antinode occurs where the standing wave vibrates at maximum amplitude Fundamental, 2nd, 3rd... Harmonics •The modes of vibration form a harmonic series •ƒ1 is the fundamental and also the first harmonic •ƒ2 is the second harmonic Example 14.9 A cello string vibrates in its fundamental mode with a frequency of 220 vibrations/s. The vibrating segment is 70.0 cm long and has a mass of 1.20 g. a) Find the tension in the string a) 163 N b) Determine the frequency of the string when it vibrates in three segments. b) 660 Hz Forced Vibrations •A system with a driving force will force a vibration at its frequency •When the frequency of the driving force equals the natural frequency of the system, the system is said to be in resonance. •This frequency is called the resonant frequency. Example 14.12 A pair of speakers separated by 1.75 m are driven by the same oscillator at a frequency of 686 Hz. An observer starts at one of the speakers and walks on a path that is perpendicular to the separation of the two speakers. (Assume vsound = 343 m/s) a) What is the position of the last intensity maximum? a) 2.81 m b) What is the position of the last intensity minimum? b) 6.00 m c) What is the position of the first intensity maximum? c) 27 cm Standing Waves in Air Columns •If one end of the air column is closed, a node must exist at this end since the movement of the air is restricted. •If the end is open, the elements of the air have complete freedom of movement and an antinode exists Harmonic Series; Tube Closed at One End Where n is odd Harmonic Series; Tube Open at Both Ends Where n is even Example 14.11 An organ pipe (open at one end and closed at the other) is designed to have a fundamental frequency of 440 Hz. Assuming the speed of sound is 343 m/s, a) What is the length of the pipe? a) 19.5 cm b) What is the frequency of the next harmonic? b) 1320 Hz