GENERAL PHYSICS PHYS 2114
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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 29 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
PHYS21 14 LECTURE 3 Computing Charged Obiect EFields Using Coulomb s Law Our goal is to find the electric field of a continuous distribution of charge We have two basic tools a The electric field of a point charge b The principle of superposition 1 Draw the picture and establish a coordinate system 2 Identify a point at which you want to calculate the electric field 3 Use any applicable symmetries to set Efield components to zero or equal to each other 4 Break up the obiect into pointlike elements 5 Write the Coulomb s Law contribution to the Efield from a representative pointlike element 6 Integrate over the entire obiect to get the Efielbleust2812008 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 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 LIZ dq S A egmenu dEx k 2 c0361 EZkQLZ rdy An lnfini re Line of Charge n Q Top view field Ex k 4 lines spread in I erz a plane lr k 2QL Side view field lines I 27quotL2 1 are parallel Eovo LIIIJE 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 IIIIIIIIIIIIIIIIIIIIIIIIIIIII 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 n 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 ie downward 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 2 Fe eE if l xl m 2m vx mi Fg mg 160x10 19 C2000 NC Z 160C2000 NC 001 m 911x10 31 kg991Nkg 2991Nkg 10 106 ms 36gtlt1013 0018m18 cm August 26 2008 Grass Seeds amp Field Patterns Field Mdp vs Field Lines Field lines start on positive charges weak The electric eld vector rs tangent to the electric eld line Field lines stop on negative charges Field vector More charge gt strong Fleld hne more field lines o o The electric field is stronger where the F39eld l39nes electric field vectors are longer and quotever quot05539 where the electric field lines are closer together nugua LU vau Problem Solving Strategy Picture Electric field lines emanate from positive charges and terminate on negative charges Solve 1 Field lines emanating from or terminating on isolated charges are drawn uniformly spaced as they emanate or terminate The number of lines emanating from a positive charge or terminating on a negative charge is proportional to the magnitude of the charge The density of lines at any point the number of lines per unit area through a surface element normal to the lines is proportional to the magnitude of the electric field there At large distances from a system of charges that has nonzero net charge the field lines are equally spaced and radial as if they were emanating from or terminating on a single point charge equal to the overall charge of the system Check Make sure that no two field lines ever intersect August 26 2008 The Dipole Field In a I b n O i I V I i K The electric eld vectors w 39393939quot n are tangent to the electric eld lines Field Lines Field Map August 26 2008 Electric Field Lines of Large and Small Distances August 26 2008 Electric field of two positive charges III a b S l 4 3 H 7 I 39 Field Mop Field Lines August 26 2008 Electric flux and field a b a c The loop is Uhquot Vector tilted by angle 0 normal to loop I I t7 Air flow The air owing through the No air ows through Vi V9959 is the component 0f the loop is maximum when 6 0 the loop when 6 90 3 VCIOCIW Perpehdlcmar 10 the 100p lmagine that you are holding the wire loop of area A in front of a fan and you want to know the volume per second I of air that flows through the loop When the loop is perpendicular to the air velocity the CD is maximal When the loop is parallel to the air velocity the CD 0 If I rotate the loop the CD is changing from CDmGX to O depends on the angle between v and vector normal to the area Electric flux A III Let us define the area vector 21 A a Area vectori is A perpendicular L0 the surface 391 he quota magnitude of A is the surface area At This definition is valid for all 3 types of surfaces including curved ones Area A Area A El Then I can define the flux of the electric field as follows CD E A this is valid for a constant electric field III For nonconstant electric field or curved surface lt13 Ed urface Question E 1 me up A A A M It A M M Plane of charge Cross section of v v v v v v v v lmxlmxlmbox E 1 NC down The total electric flux through the box is c O Nm2C b 1 Nm2C c 2 Nm2C d 4 Nm2C Gauss s law CI The electric flux Through any closed surface is proportional To The neT charge enclosed by This surface 7 i 13 ch dA80 I Here The inTegraI is Taken over any closed surface Equivalence of the Gauss s law and Coulomb s law Cross section of a Gaussian sphere of El a I39adiusrThisisamalhemalicalsurface not a physical surface radius R that surrounds a point charge q located at its center cpltfEdAEASPW 47r50R2 8 The electric field is everywhere 0 perpendicular to the surface and has Ihe sums magnitude ll everv nninl III This result is independent of how big we choose to make the spherical surface A very large or a very small surface will have the same flux which depends only on the size of the charge q at the center of the sphere Every field line passing through the smaller sphere also passes through the larger sphere Hence the ux through the two spheres is the same Question Fquot quot d fax was r t K 7 I i 2q t JCT rt t 24 ka SW 4 K 21a quot quot my 39 u v w E z k X Y What is the ranking of the electric fluxes for these systems a PX Iygt DU CDV Dw b CDXCDYCDUgt CDVCDW c PX Iygt CIquot Dv gtDw d CDV Dy gt DU 13w Dx e CDVCDYgt IUgtIwgtCIx
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