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# Calculus Based Physics II PHYS 2120

GPA 3.57

Timothy Farris

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COURSE
PROF.
Timothy Farris
TYPE
Class Notes
PAGES
30
WORDS
KARMA
25 ?

## Popular in Physics 2

This 30 page Class Notes was uploaded by Mrs. Charity Bradtke on Wednesday October 28, 2015. The Class Notes belongs to PHYS 2120 at Volunteer State Community College taught by Timothy Farris in Fall. Since its upload, it has received 13 views. For similar materials see /class/230703/phys-2120-volunteer-state-community-college in Physics 2 at Volunteer State Community College.

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Date Created: 10/28/15
253 V amp U Due to Point Charges Only the parts of the path from A to B that are parallel to E contribute to the potential Ava39g s39w A f A Equot hi r T A 54 lfs Fnrg So the potential at a distance r from a point charge q is Vr firE r w r JVC O If we bring a 2nd point charge into the vicinity of the first we can have the potential energy of the system of the two point charges V urnr l Va 139 12 U1 w t U z A gm H r Ifwe have a collection of charges the potential at any point is just the sum of the potentials of the individual charges i 13 V 2 quotz Example a Find the electric potential at the origin due to this charge configuration L 391 1 7 It I i I i W39Jx V Vi39Ht 395 Lt l kcil a 6777ko1 7h7 t 2 1 L nwu LfS39ao h LAq 3 0H1 rlera39y 77 New b Find the potential energy of the system of the 2 charges A a L m 4i r r 5 Q J 5 31 u A lt 206194335X 39 Illj 5 31 254 Obtaining the Electric Field from a Potential w 11 AV S I ll u lV quot a E 5 u mans 4 4 Exrig l J E I J 2 3 gt A lA JV E G 3 731t For example For a point charge V L zr 4v 4 J 1 E 391 1 T JI r 4 k 4 rt 75 2 E 255 V Due to a Continuous Charge Distribution M Example Find the potential at a distance quotdquot from a long line of charge of length 2L 1L Pia 139er A 3 w 91 X07 rquot 9 w L L 5 V 394 a zkf quot WW iii 1 V39 14 V lk i L V A la7 V QLALA LN73 woo V 1mg p 256 V Due to a Charged Conductor 1 In a charged conductor the potential is the same everywhere inside amp on the conductor 2 The surface of the conductor is an equipotential surface 1 E 0 inside the conductor rgt quotJ vquot 5 u A Jm E 3 V w gt 1 SJ 1 Jul1 6 39Egf AK 519 V 1 mll 39l quot Lurkt 3 The E is also zero everywhere inside a cavity in conducto I k 3901 A39 I o a This is the idea behind a Faraday cage a metal enclosure used to protect something inside from external E fields or to protect the exterior from fields inside the cage artfulquot Ch 26 Capacitance amp Dielectrics 261 Definition of Capacitance Two conductors carrying equal but opposite charges together form a capacitor We define the capacitance Q as the charge stored per unit potential Qavl Q lt 5V C 4Q S L L flhlquot A WK K We get a new unit yea r i m w 1 farad is huge We39ll usually use F tL 1 h F J39KF 9 F la12F Example If two conductors that will each hold 200 uC with a potential difference of 150 V find the capacitance r 4 rem i C inn F SW mm 242 amp 3 Gauss39 Law amp Applications Karl F Gauss recognized that the flux throughany closed surface is proportional to the charge inside amp nothing else This result is know as Gauss39 law 413 ahz Snark 4 u re re 30an 1 9 Ar 04598 5quot at A wa fq un AC1 to 117 QPf 4angA fen amprampe55 Notes Gauss39 law does notdepend on the size or shape of the gaussian surface Gauss39 law uses an quotexternalquot quantitythe fluxto probe ar quotinternalquot quantitythe net charge 4 k z Gauss39 law can be used to find electric fields if the symmetry is right Example Find the electric eld at a distance d away from a long line of charge 7 F arj z Zh quot7 me Q ah cg fa E 5 Lea 6 H 46 1 V 5 Enlw x gtAvg A 1 E 1 quot rf 921 51 244 Conductors in Electrostatic Equilibrium Electrostatic eguilibrium no net motion of charge Properties 1 E O everywhereinside a conductor 2 Any excess of charge resides on the surface of a conductor 3 E is perpendicular to the surface at the surface and equal to 6 2 49 u In P A 4 Charge accumulates at quotbendsquot amp quotcornersquot of conductors if i i Ego m 47m 1 gr an APJ J Eagfk J mowt n4 oaw nogotzkc 2 ux 797 50 Tax39039 3 475 47 39 755 int t 3ia 3b 7 17 S 19 A 4 39 E A 39 39 rA 32312 5quot 2 90w 49 a 5 7 7 5 Application of 4 lightning rods 1 7 i i 2 at 2 1 L h 1 Ch 25 Electric Potential 251 Potential Difference amp Electric Potential 5 h A Wez 5 2 9 z 39 9x A A l7 E 3 v aqua JAYS5 Much as we defined the electric field as the force per unit charge we define the potential difference as the change in potential energy per unit charge AVE 39 5 36quot We choose the reference point where U amp V 0 to be out at infinity Vrfi r a This is the potential at the point P Units De ne the voltV Wu lgwm u f LD39LLD 41 I Liftl39 Up so it s 4 39 L s391 1quot 7 M 5 73 Vn A A 50 nllt A t 1D up 0 c T A unit of energy frequently used in atomic amp nuclear physics is the electron volt eV It39s the energy an electron receives when moved through a potential difference of 1 volt l QM dd 47 linu V t I m h r 252 Potential Difference in a Uniform E Field Consider moving in a uniform electric field first moving parallel to the field 6 d a AV V3441 XAQ 2 E 393 39 lt V VA t L 0 7 2 7 Va So electric field lines point in the direction of decreasing potential The field does work when a charge move in the direction of the field lines 2 Mat40 wf x E 39 31 A UJ39 K Now moving not parallel to the field A 3 A g 7 3 l I f A gt 6 aa AW 31 3 75am YEampgtJ A B This motivates the idea of an equipotential surfacewhere the potential at every point is the same TheE is everywhere perpendicular to an equipotential surface HW Ch 24 pp 754758 Q 23 P 9 10 15 29 35 39 44 Ch25p787 P13 Io AV bv f Vl quotIWn 9v Hwa V 5 T n3 I 397 flu4 504 4 154 CH 29 Magnetic Fields 291 The Magnetic Field The magnetic field B quotcarriesquot the magnetic force Any magnetic object or any moving charge will have aB field We define the direction ofthe magnetic field to be in the direction that the quotnorthquot end of a compass poi the compass is in the fie A 5 9 v ts when w We can visualize the magnetic field with field just as we did with the electric field The relative spacing of the lines represents the strength of the B field In using the power supply you39ll have a black resistor to use instead of the lamp 354 Reflection Normal L M M Incident Re ected K EF LE 677 ray ray a 5 Example You amp a friend stand on opposite edges of a pond Your eyes are 180cm above the water level You see the reflection of his eyes at an angle of 80 below the horizontal in the pond How wide is the pond 355 Refraction Incident Normal Re ected Refraction occurs because ra ra y y a wave changes speed as A it goes from one medium 91 91 to an9ther I I I I x safer 1 I I I I I I I I I I l o J I I I I I I I I Refracted ray 2004 TTTTT o n oooooooo Ie a We find that the speed of light is different in different media We characterize the differences in speed with the index of refraction n For a particular material c quot395 ngap d It n 5 33quot ci 6quot7 A a J c gt IS in N I ix For small angles 910 81 But this is not so for larger angles For larger angles and also including small angles 963549 and the proportionality constant depends on the two materials indices of refraction The equation is Snell39s Law D I 9amp6quot F mol anal 9 o 24639 gal quot391 82 549 9 7 782 Example An aquarium with crown glass sides is filled with water What is the angle of a ray in the glass that is refracted from a ray inside the tank that strikes the glass at a 350 angle nukkr Lzs quot395 4 quot 2 Weir l Haulquot aquot 2 e 399 quotz k 39 1 115 quot7 Lu m 5239 50 quotI L33 5 9quot TE 939 mo1 Gkquot S39bi 3a 358 Total lnternal Reflection If a ray goes from an optically dense medium high quotnquot to a less dense medium lower quotnquot then 8 7539 There is a critical an le Ge for the incident ray for which the refracted ray 90 meaning there is no refraction into the 2nd material A 7 ntc n For angles 3 the critical angle there is no refraction and we have total internal reflection Vlz 9 G 1 10 quotl n A 2A8g 7 39 n2 ltn Example Dumas gives Ethyl an engagement ring Being suspicious Ethyl39s daddy rigs up a testing jig using the headlights of his John Deere and measures a critical angle of 28 in the stone of the engagement ring Is the stone diamond or cubic zirconia 1 lt 26 SM 36 gg 4 M 39r 7 quotDo I 39 Sho 39 U nl 384 AK m 139L1 A Lag 123 31 21 The phenomenon of total internal reflection is behind applications such as optical fibers HWCh 35 pp 1118 1121 P 1 6 12 13 23 37 38 Title Apr 17 932 AM 8 0f 8

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