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# GENERAL PHYSICS PHYS 2114

OK State

GPA 3.82

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This 0 page Class Notes was uploaded by Kendrick Wilderman on Sunday November 1, 2015. The Class Notes belongs to PHYS 2114 at Oklahoma State University taught by Flera Rizatdinova in Fall. Since its upload, it has received 19 views. For similar materials see /class/232918/phys-2114-oklahoma-state-university in Physics 2 at Oklahoma State University.

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Date Created: 11/01/15

LECTURE 2 General information El Website of the course is on D2L II Syllabus El Lecture schedule pdf of the lecture l Webassign is FREE for you those of you who have had problems please try again Let me know if you still have problems with login I ll put homework l tonignt on the webassign It will be due to Friday August 28 1 159 pm First quiz will be online on Monday by l 1 am You ll have 24 hours to complete it Ldst ecture we discussed I III Coulomb s law d k A F12 qlzqzr Coulomb s law 201 r III Electrostatic forces can be superimposed gt F F 152 1330nj net 1 on J on J Today s plcm II Electric field of point charge El Electric field of CI charged line Efield of CI ring see textbook II Motion in the electric field Motion and Response Ori inal F Question A g AonB B If charge A moves how long does it take for the force vector on B to a F A onB after respond by Changmg charge A moves direction 8192009 Isaac Newton vs Michael Faraday Isaac Newton Forces like gravity and electricity are examples of action at a distance The influence of A reaches across space to affect B instantly and without the need for contact Michael Faraday Forces like gravity and electricity are mediated by a field that alters the space around the mass or charge Particle A produces a field and particle B responds to this field by experiencing a force Changes in the field require time to propagate F 9 Am B We 0 Fficld on B In the Newtonian View A exerts a force directly on B In Faraday s View A alters the space around it The wavy lines are poetic license We don t know what the alteration looks like Particle B then responds to the altered space The altered space is the agent that exerts the force on B 8192009 The electric field III Define the electric field as the force per unit charge The electric eld at any point is the force per unit charge that would be experienced by a charge at that point Mathematically E electric field al m Cupyrigm 2007 Pearson Education Inc publishing as Pearson AddisonWesley The electric field permeates all of space 1 Charges interact with the electric field inducing a nonzero value of E subiect to boundary conditions III Does E depend on a the charge that detects it No because the force is proportional to q Right at this point the electric field is described by the vector If That means a point charge q placed here would experience an electric force q 1 1 4 anEz Over here7 farther llrom the charge producing the field a point charge q would experience a weaker i39nrce IE2 in a different direction The electric field is a continuous entity so there are field Vectors everywhere We just can39t draw them all I wright o 2007 u E The Field of c Poin r Charge V F A 3r field of a pomt charge H G H qtest r A Cnpyrigy u aaaaaaaaaaaa arson Edumlion m publmhing as FameI Ad lsawWaley 8182009 Question I39 gt gt 7 7 r 7 An electron q e m me is placed at the position shown by the black clot The force on the electron is CI Zero b To the right c To the left ol Insufficient information to tell 8182009 Fields of charge distributions El Electric force obeys the superposition principle so does the electric field EEIE2E3m2Ei2k Qa 204 I I l Copyright 2007 pm mm m punishing a MdimWestey Picture To calculate the resultant electric field E at field point P due to a specified distribution of point charges draw the charge configuration Include coordinate axes and the field point on the drawing Solve 1 On the drawing label the distance rip from each charge to point P Include an electric field vector EiP for the electric field at Pdue to each point charge 2 The field point P and the point charges are not on a single line then label the angle each individual electric field vector EiP makes with one of the coordinate axes 3 Calculate the components of each individual field vector EiP along the axes directions and use these to calculate the components of the resultant electric field Ep Example Calculating the Electric Field III A 10 nC charged particle is located at the origin Points 1 2 and 3 ha39ve xy coordinates in cm of 10 01 and 11 90000 MC respectively Determine the electric field E at each of these points 10 11C 90000 NC 1 cm 8182009 Question 2 In Which system has the largest electric field at the location of the numbered point I I QC 0 W A 8182009 The electric dipole II Electric dipole is the system consisting of two point charges of equal magnitude and opposite sign I Electric field of the dipole at large values of y wrt a E 2 2 y El Dipole moment pqd y gt Here s the 5 field point gt 39 E v5 E a is the x component of the displacement y r from q to the field point 2 q I t a so the x component of the unit vector from q isf39x ar cumiqu o ZCC I Peanut Educaml l luc aublislilnu as Fealson A uiwn wesl39cy and the x component of the displacement from q is a so fx ar Continuous charge distribution III If the charge distribution extends through the volume we describe it in terms of the volume charge density pCm3 I Surfoce charge density GCm2 El Lineor charge density MCm III To calculate the field of a continuous charge distribution we divide the charged region into infinitely small charge elements dq e kd A 39 E 2 J d E J qr eld ofacontmuous 207 r2 charge distribution Copyright 2007 Pearson Educa on Ina publishing as Pearson AddisonWesley Example EField of a Charged Line A thin uniformly charged rod of length L has a total charge Q Find the electric field at a distance r from the axis of the rod in the plane that bisects the rod Iina d ngdnily EQ Strategy Break up the rod into small charge elements use Coulomb s Law to calculate the E field of each and add them up What is 112 electric eld at this point quot The linear charge 0 Choose a cuurdlnnle system with demlly 15 A QL lh trigin at the entcr tithe r 1d L i L l E L12 Ey 0 Segmem39 9 Identify the point at which uc39rt s going to culchlate the eld re ectional l x T t Divide the Ind into Nsmull segments 0 ll I P gt9 3 Symme ry of length A and charge AQ AA 39 0 Draw the field vector Dismal f E Ofeling segtttcttl i about midplane 6 Nole that the field from a symmetrically located charge segment will cancel 5 L2 Example EField of 0 Charged Line 2 n Consider the x component of the electric field contribution dEx produced by a small segment i of the rod lqcated at yi with length dy Its charge dq will be QLdy 39 dq LIE Segmemi dEx kr 2c036i l kQL r LV2y2 lr2y2 k2i L r2y232 L2 E kQLf My kQ y kQ x LL2 7 2Iy232 L rwlrz y2 LZ r lrz L22 An lnfini re Line of Cha rge Q Top view field Ex 4 lines 5 read in k I erz a plan lr k 2QL ide view fie ines rl2rL2 1 sure parallel ld l Eovo tailjg 4 4 4 gt gt gt 2 4 4 4 gt gt gt litli 2 4 4 4 gt gt gt 15x 7quot Field of an infinite line of charge falls off as lr not lr2 Infinite line of charge Motion of Charges in E Fields mlIlllllllllllllllllllllllllll The motion of 5 single charged particle in cm electric field is governed by the definition of electric field and Newton s law FqE 11 a m e le Pm August 26 2008 Example Electron Moving in a Parallel Electric Field An electron with an initial velocity v0 200xl 03 ms along the x axis is proiected into a E uniform electric field E 1000 NC parallel to the x axis How far does the electron travel before being 7 brought momentarily to rest 96 I F e x ax Ex m m 2 2 vx v0x 2axAx 2 2 31 6 2 200 10 ms 114 102m ax e x gtlt ClOOO NC Example Electron Moving in a Perpendicular Electric Field An electron with an initial velocity v0 l00x10gt ms along the x axis enters a uniform electric field E 2000 NC along the y axis vo f ie downward Ye V a Compare the electric and gravitational forces on the electron b By how much is the electron deflected after travelling 10 cm in the x direction F g 36gtlt1013 6E 5 2 6El xlz z mg m 2m vx 160X10 C2000 NC 160C2000 NC 001m 911x1031 kg991Nkg 2991Nkg 10 106 ms 0018m18 cm August 26 2008

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