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UF PHY 2049 - Physics II

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PHY 2049: Physics IIquizQuiz 2Slide 4Potential Energy and PotentialSlide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25PHY 2049: Physics IIWe begin with a live clicker today.Wileyplus homework should be fully operational.Tea and Cookies: We meet on Tuesdays at 4:00PM for tea and cookies in room 2165.quizThe electric field at a distance of 1 m from an isolated point particle with a charge of 2×10−9 C is:A. 1.8N/C B. 180N/C C. 18N/CD. 1800N/C E. none of theseQuiz 2An isolated charged point particle produces an electric field with magnitude E at a point 2m away from the charge. A point at which the field magnitude is 4E is:A. 1m away from the particleB. 0.5m away from the particleC. 2m away from the particleD. 4m away from the particleE. 8m away from the particlePHY 2049: Physics IILast weekCoulomb’s law, Electric Field and Gauss’ theoremTodayElectric Potential Energy and Electric PotentialsNumerous casesPotential Energy and PotentialForce => work => change in K=> change in Potential energyΔU = Uf – Ui = -W = - ΔKWork done is path independent. K+U = constant.U = k q1 q2/r : interaction energy of two charges. Sign mattersPHY 2049: Physics IIElectric PotentialV = U/q = -W/qUnits of Joules/coulomb = volt1 eV = e x 1V = 1.6 x 10-19 JAlsoV = kq/r Vf –Vi = -∫E.dsIn case of multiple charges, add as a numberPHY 2049: Physics IICan you tell the sign of the charge by looking at its behavior over a surface of potentials. Is the speed at the end bigger or smaller than in the beginning.1. Neg., lower2. ?3. Pos., higher4. Neg., higher5. Pos., higherPHY 2049: Physics IIVf-Vi = -∫ k q/r2 drChoose Vi = V (∞)=0V(r) = kq/rV = kpcosθ/r2E = -∂V/∂s = -Uniformly charged diskV = ??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=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)PHY 2049: Physics II.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)( )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)SABVqFrErrDrConsider 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


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