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# Class Note for PH 106 at UA-Generl Physics WCalc II (3)

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Date Created: 02/06/15

UNIVERSITY OF ALABAMA Department of Physics and Astronomy PH 106 LeClair Spring 2008 Exercise Electrical Energy amp Capacitance Solutions 1 Find the equivalent capacitance of the capacitors in the gure below 3H4 SHF H H 39 2V Before We start it is useful to remember that one farad times one volt gives one coulomb lF lV l C and that capacitance times voltage gives stored charge Q0V Knowing this novv Will save some confusion on units later on For that matter it is also good to remember that the pre x 1L means 1076 In order to nd a single equivalent capacitor that could replace all ve in the diagram above We need to look for purely series and parallel combinations that can be replaced by a single capacitor The uppermost SilF and 41lF capacitors are purely in series so they can be replaced by a single equivalent We Will call 034 as shown belovv n s an r n IZV Using our rule for combining series capacitors We can nd the value of 034 easily 1 7 1 1 034 7 8 ME 8 ME 084 EMF z 267MF Now We have this equivalent capacitance purely in parallel With the second 8MP capacitor We can replace 034 and the second SilF capacitors With a single equivalent Which We Will call 0334 Cam awn 439 2V Using our addition rule for parallel capacitors We can nd its value PTPFQ s 31 ariazirriagurl sti E2 i P322 6555 2131LZuiixs iilizg iir as 1 a 3595 35 zirgxiiEranaiiiaiu avatarfzuivxyigi Engaging graiinzzwnxnlfrazwn rln 253 inwnnzw nuzsvxfuiaiwii nxuiaiaxasin1 a a awxr2ni zt2axngvm xiili Kfuz E s wmu awz WW mishan 5 m2 V s 9313 iiiri 5 Eilnzwn sffirxqi Egg L 03555 5Eir a igusvir rriizw xiux mixin vxfuai gxrz xfririarir xi 2 fa ri ir n zrpiriiawnxusxxxilfai aria 93 w 32 2350 a u s i 23 752755931zingl sizsgvxzaain a 1 05 12 z E iizsvin ign 2725 XE a A QWF 581 3 V A z 126V W 4aF 4aF QuPper 8HF 5 1MC V 7 063V upper BHF SMF SMF Now we know the charge and voltage on every single capacitance7 as well as the overall charge Q54 and effective capacitance ng Your numbers may be very slightly di erent than those above due to different choices in rounding7 this is normal The results are summarized in the table below Table 1 Equivalent aapam tances charges aaa valtages Capacitor uF Charge HO Voltage V top 8015 51 063 4M1 51 126 lower 80F 152 19 6M1 2016 34 SMF 2016 67 Ceq 2016 12 2 A parallelplate capacitor has 400 cm2 plates separated by 600 mm of air If a 120V battery is connected to this capacitor7 how much energy does it store in Joules ln electron volts If we want to nd the energy stored in the capacitor7 we need to know two of three things7 minimally the amount of charge stored7 the voltage applied7 and the capacitance Any two of these three are su icient7 based on our formula for the potential energy stored in a capacitor 1 1 2 Q2 APE 7 AV 70 AV i 2Q 2 20 We already know the applied voltage7 AV 120Volts Since this is a parallel plate capacitor and we know its area A and plate spacing d we can easily calculate the capacitance if we are very careful with units Recall that the dielectric constant of air is essentially one RR 1 K6014 d 150400cm21010 2 600 X 103 m 885 X 10 12 Fm 4 X 10quot1 m2 600 X 103 m 7 885 X 400 13 600 10 F x 590 gtlt10 13 F 0590pF C Now we know the capacitance and the voltage7 we can nd the energy readily APE 590 gtlt10 13 F 120V2 425 gtlt10 11 F V2 425 X10711J l 2 425 X 10 11J lt 16V 7 8 7 160 X1071 7 266 X 10 av 266MeV This problem brings to mind a few handy Sl unit conversions which you should be able to verify 11 1 F 1V2 1 C 1 V 1 C 1 F 1V 3 A potential difference of 100 mV exists between the outer and inner surfaces of a cell membrane The inner surface is negative relative to the outer How much work is required to move a sodium ion NaJr outside the cell from the interior Answer in electron volts and Joules A singlycharged ion has a charge of 1e 1eV16 X 10 19 J The work done in moving a charge 1 across a potential di erence AV is readily calculated W 7APE iqAV In this case the charge is e 16 X 10 19 C and AVO1V Watch how easy it is to nd the answer in electron volts 7 7 719 1eV W iqAVi 16 X 10 C 01V m WCMOJV 701 Temember1J1C1V 701 eV In fact we didnlt even need to go through all that An electron volt is de ned as the energy required to move one electron s equivalent of charge 1e through a potential di erence of 1Volt Our ion has the same magnitude of charge as an electron and we move it through 01V Following the de nition of an electron volt we must have APEO1eV This is one reason why electron volts are such a handy unit for many areas of physics Anyway how about the answer in Joules Well if 1eV16 gtlt 107191 then O1eV16 X 10 20 J 4 A point charge 4 is a distance 1 above an in nite conducting plate Given that the electric eld above the plate must be 47rkea calculate the surface charge density as a function of the position on the plate As we discussed in class a conductor acts as a mirror for electric eld lines which allows us to replace some dif cult problems involving conductors with equivalent problems involving point charges The simplest example of this is shown below a point charge just above an in nite conducting plate Our real problem involves a point charge 4 a distance I from a conducting sheet Since the conducting sheet acts like a mirror for electric eld lines we can replace the conducting plane by a second point charge 74 at a distance 21 from the original charge or a dipole This virtual charge will give eiactly the same electric eld as the induced charges in the conducting plate will the eld lines from 4 have to intersect the conducting plate perpendicular to its surface which is exactly what the eld of our dipole looks like We have already solved for the eld from a dipole completely so in fact we have solved this problem too and we know what the eld is anywhere we choose to specify We solved the virtual problem which gives the same answer as our problem What about the real charges induced in the plate We know our positive point charge will attract negative charges within the plate and draw them toward it build ing up a local region of negative charge directly below it Since we know what the electric eld is anywhere on the plate wmd mg chug more 1 mt m momma mm daemon 721mg 1mm mlrw ojmlaggs Rxght see 1m coming m zu n mu5 be the same x that ola dlpole along lts central dmdlng plane 7 we can also nd the preclse dlstrlbntlon olthls charge On very general golmds we already denved that above an ln nme condnctlng plane the electrlc eld must have a value Awkeo where o 15 the snrlace charge denslty Thus all we have to do 15 nd the electrlc eld at an arbltrary posltlon along the plate and we wlll have the snrlace charge denslty We wlll call the dstance lrorrr the posluve charge to the plate 2 rneanlng the lrrrage charge 15 a dlstance 2 below the plate and the lamal posltlon along the plane ltsell wlll be u The 91501le eld due to the real q charge 15 mlly lonnd ke l lEh 2 clearly the eld lrorrr the vrrtnal 7 charge E 15 the same In magnltnde slnce both elds must be along a hne connectlng the polnt ol lnmst on the plate wlth thelr rspectlve chargs the symmeLry ol the system dlctam that the venlcal components ol the two elds must cancel leavrng ns wlth a net eld In the 72 dlrectlon The total eld 15 then the sum olthe horlzontal components ol E and 3 whlch are ldentlcal We should not lorget the r slgn slnce we are de nlng the q charge to be a posluve dlstance 2 above the plane kg 0059 21cqu mml EM 522 mg zllzgl c059 22 W ln the lgt step we used the relatlonshlp 00592 fr 2 As reqnlred by our boundary oondlLlons lor E the eld 15 perpendlcnlar tothe plane The total eld must also be Arkpo thus 1 72m 712 Awke2zz32 2722232 And that 15 rt the snrlace charge denslty g a lnnctlon olthe lamal posltlon on the plate 7 and the dlstance ol the q when nears we are xaa y seam up an as aylmdrmal meme systenn you are mtexestsd m suahthmg charge from the plate x The surface charge density is really only a function of r as it exists only on the surface of the plate I is in this case basically a parameter that characterizes the strength of the polarizing q charge For an actual dipole rather than a virtual one the product of the charge and separation 211 in this case is known as the dipole moment and it really is a vector measure of a dipole7s e ective strength This result makes some sens out by the lack of 9 dependence in our nal result Below we plot a crossesection of the surface charge density along the plate for various values of x in the plane of the plate the charge density must be radially symmetric and this is borne rdltp d39r auteee ehuge aeaaty Figure 2 Left Surfa4a 4haw ga danslty as a fun4tlsn sf tbs tatsVal paaltma an tbs plats fsr tempts distan4ss ppm tbs plats Right Elsmsnt sf aV Ea m polar 4ssrdlnatss There is of course one more check we can make If our method of images is correct then all of the surface charge density over the entire plate must just add up to aq the same as the image charge we placed11 Within the conducting plane we can integrate 042A over the whole plate to nd the total charge Since the charge density is radially symmetric it makes sense to integrate over area in polar coordinates a let the radial distance vary from Zero to in nity and sweep the radius through an ineplane angle p An element of area in polar coordinates is r dr d9 you should be able to see how this comes about from the gure above We will integrate over all possible patches like this on the sheet which means taking r from o a gt0 and p from o a 27r The integration over p is trivial since nothing depends on the angle and it just adds a factor 27r 2quot ae ae iqxr arr aquot a gene fadA 41W farm dc2 f mdr aqf mda o 0 0 0 27r r2 x2 r2 x2 0 q7 25Qq 2 c5q071gc 7391 No problem There axe deep aha ecmpiieatea theorem that Prove that this must be true We will hat wcny about the general eaae but juat peeve one one aciutich ia aehaibie

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