PHY 2049: Physics IISlide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25PHY 2049: Physics IITutoring Center is open in room NPB 1215, M-F 12:00AM -4:00PM. It is free.Hopefully all homework problems have beensolved. Please see me immediately after the class if there is still an issue.PHY 2049: Physics IIAbout bicycle riding: If some one shows you a trick, it means that,It is possible to do that trickThe person who showed you, can do the trick. It does not mean that you can do the trick.Try it a lot of times and you can too.PHY 2049: Physics IIForce => work => change in K=> change in Potential energyΔU = Uf – Ui = -W = - ΔKWork done is path independent. K+U = constant.U = k q1 q2/rPHY 2049: Physics IIElectric PotentialV = U/q = -W/qUnits of Joules/coulomb = volt1 eV = e x 1V = 1.6 x 10-19 JAlsoV = kq/r Vf –Vi = -∫E.dsAdd as a numberPHY 2049: Physics IIPHY 2049: Physics IIRank them by the magnitude of electric fieldWhich has the field pointed to the rightWhat does an electron do when released midwayPHY 2049: Physics IIPHY 2049: Physics IIVf-Vi = -∫ k q/r2 drChoose Vi = V (∞)=0V(r) = kq/rPHY 2049: Physics IIV = kpcosθ/r2E = -∂V/∂s = -Uniformly charged diskV = ??ABCprConsider the electric dipole shown in the figureWe will determine the electric potential created at point Pby the two charges of the dipole using superposVExample : Potential due to an electric dipole( ) ( )( ) ( )( ) ( ) ( ) ( )ition.Point P is at a distance from the center O of the dipole.Line OP makes an angle with the dipole axis 14 4We assume that where is o orr rq q qV V Vr r r rr d dqpe pe- ++ -+ - - +� �-= + = - =� �� �� �?( ) ( )2( ) ( )2 2the charge separationFrom triangle ABC we have: coscos 1 cosAlso: 4 4where the electric dipole momento or r dq d pr r r Vr rp qdqq qpe pe- +- +- �� � � == =(24 - 6)21 cos4opVrqpe=.PrdqConsider the charge distribution shown in the figureIn order to determine the electric potential createdby the distribution at point P we use the prVPotential due to a continuous charge distribution :inciple ofsuperposition as follows:We divide the distribution into elements of charge For a volume charge distribution For a surface charge distribution For a linear charge distributidqdq dVdq dArs==1. on dq dl= l1 We determine the potential created by at P 41 We sum all the contributions in the form of the integral: 4 The integral is taken over the whole charge distriboodqdV dq dVrdqVrpepe==�2.3.Note 1: ution The integral involves only scalar quantities Note 2 :14odqVrpe=�(24 - 7)dqOAPotential created by a line of charge of length L and uniform linear charge density λ at point P. Consider the charge element at point A, adistance from O. From triangle OAP wedq dxxl=Example : ()()()2 22 22 202 22202222 have: Here is the distance OPThe potential created by at P is:1 14 44ln4lln ln4no oLoLoor d x ddV dqdq dxdVrd xdxVd xV x d xV L L xdxxdx dd xlpe pelpelpelpe= += =+=+� �= + +� �� �� �= +=+ -� �� �+ ++��(24 - 8)( )F+r( )F-r22 2Many molecules such as H O have a permanent electricdipole moment. These are known as "polar" molecules. Others, such as O , N , etc the electric dipole moment is zero. These arInduced dipole momente known as "nonpolar" moleculesOne such molecule is shown in fig.a. The electric dipolemoment is zero because the center of the positivecharge coincides with the center of the negative charge.In fprig.b we show what happens when an electric field is applied on a nonpolar molecule. The electric forces on the positive and nagative charges are equal in magnitudebut opposite in directionErAs a result the centers of the positve and negative charges move in oppositedirections and do not coincide. Thus a non-zero electric dipole moment appears. This is known as "induced" electric dipolpre moment and the moleculeis said to be "polarized". When the electric field is removed disappearspr(24 - 9)A collection of points that have the samepotential is known as an equipotential surface. Four such surfaces are shown inthe figure. The work done by as it movesa charge beEqEquipotential surfacesrtween two points that have apotential difference is given by: VW q VD=- D( )( )( )2 12 1For path I : 0 because 0For path II: 0 because 0For path III: For path IV: When a charge is moved on an equipotential surface 0The work donIIIIIIIVW VW VW q V q V VW q V q V VV= D == D == D = -= D = -D =Note :e by the electric field is zero: 0W = W q V=- D(24 - 10)SABVqFrErrDrConsider the equipotential surface at potential . A charge is moved by an electric field from point Ato point B along a pathV qEThe electric field E is perpendicular to the equipotential surfacesrr . Points A and B and the path lie on S rDrLets assume that the electric field forms an angle with the path .The work done by the electric field is: cos cosWe also know that 0. Thus: cos 00, 0, E rW F r F r qE rW qE rq E rqq qqD= �D = D = D= D =� � D �rrrr0 Thus: cos 0 The correct picture is shown in the figure belo9w0qq = � = �(24 - 11)SVErExamples of equipotential surfaces and the corresponding electric field lines Uniform electric fieldIsolated point charge Electric dipole(24 - 12) Assume that is constant constant4 4Thus the equiptential surfaces are spheres with their center at the point chargeand radius 4o ooq qV V rr Vqrpe pepe= � = == Equipotential surfaces for a point charge q :VABVV+dV Now we will tackle the reverse problem i.e. determine if we know . Consider two equipotential surfaces that corrspond to the values and E VE VV V +Calculating the electric field from the potential rr separated by a distance as shown in the figure. Consider an arbitrary directionrepresented by the vector . We will allow the electric field to move a charge from the equipotenbtial surfodVdsdsqrace to the surface V V dV+The work done by the electric field is given by: (eqs.1) also cos cos (eqs.2)If we compare these two equations we have:cos cosFrom triangle PAB we see thatooo oW q dVW Fds Eq dsdVEq ds q dV Edsq qq q=-= ==- � =- cos is the component of along the direction s.Thus: ssEE EVEsq�=-�r sVEs�=-�(24 -
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