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# Phys 272 Exam 1 Study Guide!! PHYS 27200 - 001

Purdue

GPA 4.0

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This 9 page Study Guide was uploaded by Alan Hwu on Friday October 2, 2015. The Study Guide belongs to PHYS 27200 - 001 at Purdue University taught by Erica Carlson in Summer 2015. Since its upload, it has received 347 views.

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Date Created: 10/02/15

Alan s Phys 272 Exam 1 Study Guide Foreword Dear all 272 fellow Thanks a lot for purchasing my first study guide I hope this can help a lot for your studying The following notes are the types of problems and quick notes that I found show up a lot in quizzes homework and past exams My guide break down into 4 sections 1 Chapters Covered 2 Essential Formulas 3 Sides notes 4 Contact info I personally found this type of organization in learning the most beneficial I highly suggest to practice past exam to get a feel of it When reading my study guide pay close attention with every symbols Typically parentheses and brackets ltgt are for showing the unit of each constant in the formula and addition notes of the content Onestar stands for PAY ATTENTION to this special case and double star means YOU MUST KNOW THIS when doing the exam Thanks again for subscribing I hope everyone get good gradeI I left my contact information below feel free to send my email for any concerns and questions Happy Studying Alan Chapters Covered Chapter 13 Electric Field Chapter 14 Polarization Insulators amp Conductors Field of a Rod Chapter 15 Field of Ring Disks Capacitor Chapter 16 Potential Potential Energy Essential Formula ltNot all of the formula listed below need to be memorized I attached a formula sheet that is gonna be given on the test Learned Smartgt Chapter 13 Coulomb force law q charge r radius E0 8854e12 unit Newton N l O O E Fq q charge unit coulomb C 1 Q E 4778 7 0 Q is the charge at the source point Question type 1 v 39LHJN 391 39 The net electric field at observation point ltx00gt E 1 ZqTS quot111 4355750 322 r 52 If r gtgt 5 if r is 100 times greater than S 1 23939 3 Ll lili 433 r Simplified formula Chapter 14 Shape of Dipole Electric Field 1 Along with the dipole axis a 1 gsqh E 33 4ne T ltThe unit vector at the end depend on the direction of the dipolegt gt H F v z I I 2 Perpendicular to the dipole 1i 5 E qt 4 e r3quot ltThe unit vector at the end depend on the direction of the dipolegt ltUse this equation When the asked point is pezpezzdz39cular to the dipolegt Chapter 15 Electric field of a Rod infinite Rod Electric constant Finite Rod of a bisecting plane 1 a I 1 4mg Is22 g Infinite Rod of a bisecting plane E f 1 QlQLi 4mg 3 quot Electric field of a ring Eton 1 Q3 waning Electric field of a disk s 1 Q Em 1 23 2E9 WEE Infinite plane E 263 Electric field of hollow spherical shell of charge Inside E 0 E 2 1 if 47mg lirl2 Outside r observation point to center of sphere Electric field of solid sphere of charge ER 2 Q L3 Inside 4mm R r observation point to center of sphere R radius of the sphere 1 1 7 2 39A 47115 li39r 2 T Outside r observation point to center of sphere Chapter 16 Electric Potential AV Uniform field Nonuniform field Fairly important to Hand Written part 1 v 39 Vr 4 2 Potential of Point charge EH lquot distance q point charge r Electric Potential Energy I QEA E qV deltaV difference in potential q charge E electric field deltaX displacement Electric Density of Electric Field V N U electric potential energy V potential E0 885e12 Etot total electric field Dielectric Constant net Total Electric field with dielectric medium 3quot Eapplied Enet total electric field Eapplied initial applied E on dielectric medium K dielectric constant ltDielectric medium will dampen the electric field The dielectric constant K varies depend on the materialsgt Side notes 1 0 Electric constant 4mg 91OA9 ltmemorizing this rather than actually plug in E0 will save you a lot of time when solving problemgt 0 Electric Potential Problem Opposite direction movement from electric field will yield positive Potential difference Potential increase Likewise moving in same direction of electric field will have yield negative potential difference Potential decrease Electric potential is result from dot product of electric field and direction of displacement therefore perpendicular movement to electric field will have no change in electric potential Diagonal movement problem only need to calculate the parallel component electric potential difference When writing hand written problem try to use the integral formula to solve problems 0 Electric field due to a RodRingDisk problem Always identify the shape first obviously in order to decide the formula use Pay close attention to the difference between variables 0 Under several condition some equation can be simplified and easy to compute 0 Example Infinity Rod and Infinity plane Some type of question have the observation point inside a charged sphere Q If you want to calculate the electric field due to the rod outside the sphere you can just simply ignore the charge sphere to do the calculations 0 The formula state that E inside the sphere is zero However that is the E from the sphere not including other surrounding objects therefore should also consider the electric field from other objects 0 Dielectric Constant I This type of question will normally want you to calculate the final potential after placing a dielectric medium between two charged plates l 1 After know the original electric field E between the plates divide it by the dielectric constant K to get the electric field after dampen ltA easier way to remember the formula the medium is damping the electric field so you have E over K I 2 Subtract the original gap by the width of the medium and divide by 2 to obtain the new gap between each plate with the medium I 3 To obtain newgap voltage take original E multiply by new gap between plate and medium I 4 Electric potential across medium is the new electric field after damping multiply the width of the medium I 5 Sum the electric potential across medium with 2 of the new gap voltage each plate has a gap with medium to get the total Potential difference between 2 plates with medium Contact info Feel free to contact me if you have any confusions or questions with my guide and materials Purdue Email thupurdueedu Phone 2246336685 LineID alanhu1218 add me Play hard Study smart Equations 5 point charge a 1 1 AV L point charge 471 60 ff I 1 qr AUG AV qAV Abs 1 d AV f Eodl AV z ZExm EyAy 5132 Speci c Results Electric field due to uniformly charged spherical shell outside like point charge inside zero 1 Q E 139 er endicular from center I rod 4760 r39r2 p l o 1 2Q L 1 q od z 4n601 1f39rltlt L ringl 2 along ans QA z I I QA 5 QA EM 260 1 2 R2 4 along am Ed 260 1 R 260 1r ltlt I A A Empamor x Q Q or Q dialog E fringel z 2 just outside capacitor 60 f0 1 213 I lEdipd u z along dipole ans where r gt s l 0 1 V labpd ch z along axis perpendicular to dipole axis where r gt s39 quot 71 60 1 Electric dipole moment pqs Polarization 17 if l l vqs i 114132 I q quotAF quot uE Ea licd 9 Edielectric p 39 kinetic energv z mv if u ltlt c Geometry 9 area of Circle 7nquot Circumference of arch 27rr area of curved surface of cylinder 2m39L surface area of sphere 4 1 2 l 393 volume of spher L imquot arc length 1130

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