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Date Created: 09/23/15
Chapter 20 Electric Potential and Capacitance 1 The concepts covered in this chapter are 1 Electric potential 2 Electric potential energy 3 Capacitance 4 Capacitors arranged in series and parallel con guration 5 Air and dielectric lled capacitors 6 Potential energy stored in capacitors 2 Potential and Potential Energy In the previous chapter we learned that charges exert a force on each other according to Coulomb s law This means that such forces that occur among the charges can change a given con guration of charges in the eld view of electrostatic interaction the electric elds of various charges interact to make the charges move ie the electric eld does work on the charge con guration The con guration can also be changed by an external agent ie work can be performed by an external agent This change in charge con guration will show up as a change in the potential energy of the system the charge con guration The change in the potential energy between or of two charges is de ned as Aw eldAWext AU 1 l7 l7 AUabUa7Ub FEWIEE 2 The potential difference AV Vb Va is de ned as b A 4 vb va JEdr 3 Potential can be de ned as the potential energy per unit charge recall a similar de nition of the electric eld E as the force per unit charge qE F A point charge Q creates an electric potential Vr at a distance r from it given by Vr kQr referenced to in nity ie Voo 0 4 The net potential at a given point due to a collection of charges is simply the algebraic sum of individual potentials Vk 5 11 1 Potential energy of two point charges is given by U12 kQJQzV 6 Potential energy of a collection or con guration of point charges is the algebraic sum of the potential energy of all the pairs of charges you can form in the con guration For example in Solved Example 1 below U is the potential energy of the 3charge con guration Potential energy of a single charge in a collection or con guration of point charges is the algebraic sum of the potential energy of all the pairs the given charge forms in the con guration For example Solved Example 1 the potential energy of Qlin part a would be the sum U12 U13 Points to remember 1 Make sure you understand the difference between potential and potential energy and between potential energy of a charge con guration and the potential energy of a charge in a charge con guration 2 A single charge will create a potential at a point in space but you need at least two charges for the potential energy to have any meaning 3 Potential and potential energy are scalars 4 Eq 3 can also be expressed as Ex dVdX and similar relations in the y and zdirections This allows one to calculate the components of E a vector by knowing the potential VXyz which is a scalar Please see sec 204 of your Text for additional details 3 Capacitance The capacitance C is de ned as C QV in the text C QAV C is always positive When we talk about the charge on a capacitor we mean the magnitude of the charge on one of its plates The capacitors can be arranged in a series or a parallel arrangement In a series arrangement i the net capacitance C is given by JC 1C11C21C3 7 ii the magnitude of the charge on every capacitor is the same iii the potential difference across each capacitor is different and is given by V QCi In a parallel arrangement i the net capacitance C is given by CC1C2C3 8 ii the potential difference Vacross each capacitor is the same iii the charge on every capacitor is different and is given by Q VCi The capacitance of a parallel plate capacitor of plate area A and separation d is given by CA80d 9 The energy stored in a capacitor is given by U 12CV2 10 By using C QV eq 10 can be cast in terms of other parameters of interest Insertion of a dielectric medium in a capacitor will change its capacitance by a factor of K where K is the dielectric constant of the medium C K KCg 4 Solved Examples 1 Three charges are arranged at Y the vertices of a rightangle Q1 triangle as shown with Q 4 uC 2Q2 and Q3 3uC a Determine the total potential energy of the threecharge system b Determine the work done by an eXternal force to move Q2 from its present position to in nity Q3 53 Q2 v X 500m gt a The initial potential energy 12 X 103912 6 X 103912 8gtlt103912 3gtlt10 2 5gtlt10 2 4gtlt10 2 U1 9gtlt109 288J b after Q2 has been removed to in nity the potential energy of the remaining charges is 12 X 10 12 the work done by the external force is given by AW AU Uf U 3607 288 0 72 Uf9gtlt109 360J Note in the problem above we have used AW AU Uf U where AU is the change in the potential energy of the charge configuration We could have also done the problem by looking at it from the point of view of Q2 in a manner of speaking charges don t really have a point of view In this case AU would be the change in the potential energy of Q2 AU UQz UQ2 07 U12 U23 9109 12 2010 0 721 2 Prob 2018 from the text discussed in lecture This is a rather longish method but has several explanatory details The problem can also be solved in a much more compact way The shorter method follows the long one a In an empty universe the 20nC charge can be placed at its location with no energy investment At a distance of4 cm it creates a potential 450 kV V 7 kgql 9x109 NmZc220x10 9 c 1 r 004 In To place the 10nC charge there we must put in energy U12 1le 10x10 9 c45 gtlt103 V 450gtlt10 5 J Next to bring up the 20nC charge requires energy U23 U13 113V q3V1 q V2 W 79 79 QOXlore C9X109 Nmzczmwj 004m 008 m 7450gtlt10 5 J450gtlt10 5 J The total energy of the three charges is U12U23U13 7450gtlt10 J b The three xed charges create this potential at the location where the fourth is released 20gtlt10 9 10gtlt10 9 7 20gtlt10 9 0042 0032 003 005 VV1V2V39x109 NmZC2 jCm V 300x103 V Energy of the system of four charged objects is conserved as the fourth charge ies away 1 2 1 2 mv qV mv qV 2 I 2 f l 79 3 7 713 2 040gtlt10 CX300gtlt10 V 2200gtlt10 kgv 0 2l20gtlt10 4 J 4 v 13 346gtlt10 ms 2gtlt10 kg The shorter method a U U12 U23 U13 U13 9109 400103918z10392 45 10395 J why is U12 U23 0 b U KE i U KE f conservation of ener from last term gy Ui KEf Ui U14 U34 U24 U24 9109 400103918310392 120 10395 J 122103913v2 why is U14 U34 0 v 12 2104ms 346104ms 3 Consider the three capacitors connected as shown A potential difference of Vac I20Vis maintained between points a and c 4H1 a What is the potential difference V55 across the 2uF capacitor b Determine the charge on the 6 4M capacitor c How much energy is stored in 2HF the 6M capacitor 6 HF 21 The 4 LLF and the 6uF capacitors can be combined into a single equivalent capacitor of capacitance I OyF Vab QIOLF V55 Q2M or Vat Vbc 210 15 5 Vab Vb Also Vab V55 I20V solving the last two equations gives Vab20V and V55 100V h Q4 an4103958105c c U 5105 m 121051 4 prob 41 solved in lecture b c 250 4F c 50500 850 4F quoti 850 4F 200 4F 7 50 quotP I c 600 200 0 up Q CAV 596 yF150 V m 200 AVQM447V 0r m c 200 4F 1507 447 1053 V FIG 112041 Q CAV 500 yF1053 V 8957 632 on 150 AF and 300 AF Concept Questions 1 A spherical balloon contains a positively charged object at its center i As the balloon is in ated to a greater volume While the charged object remains at the center does the electric potential at the surface of the balloon a increase b decrease or c remain the same ii Does the electric ux through the surface of the balloon a increase b decrease or c remain the same Answer i b The electric potential is inversely proportional to the radius see Eq 2011 ii c Because the same number of eld lines passes through a closed surface of any shape or size the electric ux through the surface remains con ant Vki Eq2011 r 2 In Active Figure 206a take qr to be anegative source charge and qz to be the test charge i If qz is initially positive and is replaced With a charge of the same magnitude but negative does the potential at the position of qz due to ql a increase b decrease or c remain the same ii When qz is changed from positive to negative does the potential energy of the twocharge system a increase b decrease or c remain the same I Active Figure 206 a Lftwo point charges are separated by a distance 2 the potential energy ofthe pair of charges is given by lamam b If charge qr is removed a potential quzriz exists at point P due to charge q Answer i c The potential is established only by the source charge and is independent of the test charge ii a The potential energy of the twocharge system is initially negative due to the products of charges of opposite sign in Equation 2013 When the sign of q is changed both charges are negative and the potential energy of the system is positive U qle k M Eq 2013 m 3 i In a certain region of space the electric potential is zero everywhere along the x aXis From this information we can conclude that the x component of the electric eld in this region is a zero b in the x direction or c in the 7x direction ii In a certain region of space the electric eld is zero From this information we can conclude that the electric potential in this region is a zero b constant c positive or 1 negative Answer i a If the potential is constant zero in this case its derivative along this direction is zero ii b If the electric eld is zero there is no change in the electric potential and it must be constant This constant value could be zero but it does not have to be zero 4 A capacitor stores charge Q at a potential difference AV If the voltage applied by a battery to the capacitor is doubled to 2 AV a the capacitance falls to half its initial value and the charge remains the same b the capacitance and the charge both fall to half their initial values c the capacitance and the charge both double or d the capacitance remains the same and the charge doubles Answer d The capacitance is a property of the physical system and does not vary with applied voltage According to Equation 2019 if the voltage is doubled the charge is doubled C 5 Eq 2019 2 AV 5 Two capacitors are identical They can be connected in series or in parallel i If you want the smallest equivalent capacitance for the combination a do you connect them in series b do you connect them in parallel or e do the combinations have the same capacitance ii Each capacitor is charged to a voltage of 10 V If you want the largest combined potential difference across the combination a do you connect them in series b do you connect them in parallel or e do the combinations have the same potential difference Answer i a When connecting capacitors in series the inverse of the capacitances add resulting in a smaller overall equivalent capacitance ii a When capacitors are connected in series the voltages add for a total of 20 V in this case If they are combined in parallel the voltage across the combination is still 10 V 6 You have three capacitors and a battery In which of the following combinations of the three capacitors will the maximum possible energy be stored when the combination is attached to the battery a When in series the maximum amount is stored b When parallel the maximum amount is stored c Both combinations will store the same amount of energy Answer b For a given voltage the energy stored in a capacitor is proportional to C U CA V2 2 Therefore you want to maximize the equivalent capacitance You do so by connecting the three capacitors in parallel so that the capacitances add CHAPTER7 22 MAGNETIC FORCES AND MAGNETIC FJELDS A statlonary charge q expenences aforce Fm ah elecm eldE accordmg to F 4111 A charge q momhg wlth veloclty v expenences a force F lh a magmth fleld B accordmg to F qvxK The magneth eld B ls measured lh uhlls ofTesla r The dlrecuon ofF eah be determined by rst hdhg the dlrecuon 0mg uslng the nghtr aposltlve charge and opposlte to that of m for a hegahve charge mmquot kxk 0 See solved Examplea below 1 A l how to deal wth sueh a Cn cularmohon see section 22 3 formers detalls You should also read the sections that deal with 959 force on a current carrying wire You should know how the equation describing the force F ILXB can be used to nd the torque on a currentcarrying loop in a magnetic eld BiotSavart law Ampere s law learn how to determine the B eld for a long wire Force between two current carrying wires How to calculate the net B eld produced by several currentcarrying conductorssee solved Example3 below A FEW SOLVED EXAMPLES 1 A particle of mass m 1030kg and charge q32x 1019C of unknown polarity moving with a velocity of I 06ms enters a region to the right of line AC as shown A in the accompanying diagram of uniform magnetic eld B of magnitude 1 0 0G The B eld points out of the page and is perpendicular to the velocity of the charge After entering the B eld the charge moves in a circular trajectory as shown a Determine the radius of the circular trajectory qu mv2r or r mvqB 1030106 32x019x103 33x103m b What is the polarity or of the charge Ans The charge is positive C c How much work was done on the particle by the magnetic eld as the particle traveled the half circle Ans Zero The magnetic force and the displacement are perpendicular to each other 2 a A proton of mass M and charge Q 16x1019C is traveling with uniform velocity V 2x1 05 k ms under the combined in uence of a magnetic eld B 06i 05j T and a uniform E eld Determine the direction and magnitude of E Solution Fm qva E 0 E va va 2x10506i 05jxk 2x105 06j 051 E 2x105062 05212 156x105NC 9 Ian 165 500 500 below the XaXis 2b A wire carrying a current 30A lies in the xyplane as shown The linear mass density of the wire is A 15x102 kgm Determine the direction you can use the diagram to the right to show the direction and magnitude of the smallest B eld required to levitate the wire in air Solution LXB B 900 L mg The magnetic eld B is perpendicular to L and in the xyplane ILB MgB 1g115x102x103 5x102T 3 Two long parallel wires oriented along the normal to the plane of the page and carrying a current I 150A are separated by d 20 cm a Determine the magnitude and the direction of the net B eld at point P P is in the plane of the paper Note you may nd it helpful to indicate the direction of the B eld produced by each wire and the direction of the net B eld on a clearly labeled diagram 132 Bi 3 on27z2102 15104T using Ampere s Law ET ZBcos30i 26104ir P 30 E y 300 ET 1 B1 0 b If an electron passes through P with a velocity 2 5x1 0 ms parallel to the two wires and directed into the page determine the magnitude 2 and direction of the force on the electron 1 d 1 eqx X J J F VB 1610 25103k 261039439 10410 19 N Solution to Lecture Problems 141 W up 1 out at the page smce the hang IS negative a nu de eiuun d into the page 7 P2213 5 mfll 682 and R t 100 Ignoring Ielnlivisuc cunecuon the kinetm enegi at Lhe elecuons 15 2qu 133gtlt10S mm m meNewton s secund law m annu n kg133gtlt105 ms MR 1170x104 Cl08gtlt10 m FIG F2213 P2227 Mal 39PZZ22 tar rm 80m39l Mom In 004 mums NA m13m l quot mono N m w rm gm www40wa N mtSoDO revmural Jud n E 1 rev o0 15 In one lmlr I39EVQIUUDU the work IS WUnm rum ABcos isuzwscom39 2125 JNIAB lloAD X 10 N m 128 x 103 T Lu one mu revolution w 21113 x 10quot1 1 w 75a gt 10 5 id 3 A lo 5 The peak power in b 1 g l EAtel by the factor ANSWERS T0 QUESTIONS This section is good for clarifying concepts introduced in this chapter The answers are given below For the statements of questions please consult your textbook Q2211fthey 39 quot L quot 39 39 39fieldL L opposite sign Q222 Similarities Both can alter the velocity ofa charged particle moving through the eld Both exert forces directly proportional to the charge ofthe particle feeling the force Positive and negative charges feel forces in opposite directions Differences The direction of the electric force is parallel or antiparallel to the direction ofthe electric field but the direction ofthe magnetic ft Femic forces can accelerate a charged particle from rest or stop a moving particle but magnetic forces canno Q223a The q gtlt E force on each electron is down Since electrons are negative V x E must be up With V to the right TB must be into the page away from you bReversing the current in the coils would reverse the direction of making it toward you Then V x E is in the direction right gtlt toward you 2 down and q x 13 will make the electron beam curve up Q224 If the current is in a direction parallel or antz39parallel to the magnetic field then there is no force Q225 Yes If the magnetic field is perpendicular to the plane of the loop then it exerts no torque on the loop Q226 If you can hook a spring balance to the particle and measure the force on it in a known electric field then q 2 will tell you its charge You cannot hook a spring balance to an electron Measuring the acceleration of small particles by observing their de ection in known electric and magnetic fields can tell you the chargetomass ratio but not separately the charge or mass Both an acceleration produced by an electric field and an acceleration caused by a magnetic field depend on the properties of the particle only by being proportional to the ratio 1 m Q228 The spiral tracks are left by charged particles gradually losing kinetic energy A straight path might be left by an uncharged particle that managed to leave a trail of bubbles or it might be the imperceptibly curving track of a very fast charged particle or of a very massive charged particle Q229 The magnetic field created by wire 1 at the position of wire 2 is into the paper Hence the magnetic force on wire 2 is in direction down X into the paper 2 to the right away from wire 1 Now wire 2 creates a magnetic field into the page at the location of wire 1 so wire 1 feels force up X into the paper 2 left away from wire 2 FIG Q229 Q 210 No total force but a torque Let Wire 1 carry current in the y direction toward the top ofthe page Let wire 2 be a millimeter above L r39 am an 39 39 39 39 On 39 r r 1 wire 1 direction which exerts force in the ix 1 righthand side wire 1 produces magneticiieldin the 71 direction and makes a 3x612 4 force of equal magnitude act on wire 2 waire 2 is 4 direction on wire 2 On the fr t move its center section will twist counterclockwise and then be M v W attracted to wire 1 r Egan FIG Q2210 MULTIPLECHOICE QUESTIONS 1 39 39 r r r L Amagnetic a a L L I of upward out ofthe page 139 into the page Answer e 39 L 4 the he an e charge on the electron 2 m A 39 39 39 39 39 39 circle with a m S V w I I I I radius ofthe circular path is the radius ofthe new path a smaller b larger or c equal in size ii An identical particle enters the field with a perpendicular to l but V the with a higher p a smaller b larger or c equal in size Answer i a The magnetic force on the particle increases in proportion to B The result is a smaller radius as we can see from Equation 223 r mv qB ii b The 39 39 39 39 39 L L 39r acceleratjon as we can see from Equation 223 3A 39 39 39 39rr r r Thewire Lg of downwardinto the page Answer c 4 Consider the current in the length of wire shown in Figure 2224 Rank the points A B and C in terms of magnitude of the magnetic eld due to the current in the length element d39s39 shown from greatest to least 5 If gt 4 t gt I Figure 2224 Where is the magnetic eld the greatest Answer B C A Point B is closest to the current element Point C is farther away and the eld is further reduced by the sin 0 factor in the cross product as X f The eld atA is zero because 0 0 5 A loose spiral spring is hung from the ceiling and a large current is sent through it Do the coils a move closer together b move farther apart or c not move at all Answer a The coils act like wires carrying parallel currents and hence attract one another
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