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Honors Physics Laboratory for Engineers and Scientists II

by: Elmira Macejkovic

Honors Physics Laboratory for Engineers and Scientists II PHY 2049H

Marketplace > University of Central Florida > Physics 2 > PHY 2049H > Honors Physics Laboratory for Engineers and Scientists II
Elmira Macejkovic
University of Central Florida
GPA 3.56

Sergey Stolbov

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Sergey Stolbov
Class Notes
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This 30 page Class Notes was uploaded by Elmira Macejkovic on Thursday October 22, 2015. The Class Notes belongs to PHY 2049H at University of Central Florida taught by Sergey Stolbov in Fall. Since its upload, it has received 49 views. For similar materials see /class/227626/phy-2049h-university-of-central-florida in Physics 2 at University of Central Florida.

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Date Created: 10/22/15
Lecture 14 Review of the chapters 23 through 27 Coulomb s Law q1 12 E12 t 12 12 F kqqu 1 12 2 12 Force on2due tol r12 k 47c80391 90 x 109 NmZC2 80 permitivity of free space 886 X 1039L2 CZNm2 Coulomb s law describes the interaction between bodies due to their charges Superposition Principle the total force on the test charge equal to the sum of the forces exerted by individual source charges F21y q3 I Forces add vectorially I IEF21xquotquot31x quot39FZIV39I39F31vM The Electric Field Group of fixed source charges oexert a force F given by Coulomb s law on a test charge qtest at position qtest The electric eld E at a given point in space is the force per unit charge that would be experienced by a test charge at that point E F qtest This is a vector function of position Electric Fields Produced by Continuous Distributions of Charge For discrete point charges we can use the superposition principle and sum the fields due to each point charge qz Electric field by C14 The electric ux I of the electric eld E through the at surface A as I E A Where A is a vector normal to the surface magnitude A and direction normal to the surface outwards in a closed surface area A quotquotquotquotquotquotquot w F is a scalar product of two vectors F E A E A cos q For an arbitrary curved surface CD I I E For a closed surface in external field no source of field inside the surface the number of incoming lines equal to the CD I I E 0 number of outgoing lines total flux 0 A charge source of electric eld is inside a closed surface N o incoming lines all lines are outgoing ux 0 Change shape or size of the closed surface the number of outgoing eld lines is not changed ux is the same Gauss s Law The electric Z q ux through any closed inside surface equals to enclosed I gtqu ZJ DE 0 A charge 80 8 0 Electric eld produced I E dA Q 80 byalx ntcharge IEdAEIdAEA A 475r2 EAE 471 2Q80 1 472190 r k 1 4 1 80 80 PermittiVity Coulomb s Law I 80 885X103912 CZNm2 Problem Sphere of Charge Q Find Er at a point outside the charged sphere Apply Gauss s law using as the Gaussian surface the sphere of radius r pictured E dA Exactly as the point charge Q is at the origin for rgtR What is the enclosed charge Q What is the flux through this surface o Eaamp EaA E dA EA E47zr2 Gauss CI Q 80 Qgo c1gt E47rr2 So 17117 1 92 9 471190 r Conductors In conductors some electrons are not bound to atoms or chemical bonds between atoms These electrons are free to move What is an electric eld inside a conductor Consider a conducting slab in an if E external static eld gt gt If the field 1ns1de the conductor were not zero free electrons in the conductor u u would experience an electrical force and thus accelerate the electrons redistribute until the magnitude of the internal eld equals the magnitude of the external eld The net eld is always zero inside the conductor What is electric charge inside a conductor Choose a gaussian surface inside but close to Gaussian the actual surface surface The electric eld inside is zero 39 Therefore net ux through the gaussian surface is zero Gauss s law If the ux 0 no charge inside Because the gaussian surface can be as close to the actual surface as desired there can be no charge inside the surface Since no net charge can be inside the surface any net charge must reside on the surface If a conductor is charged surface charge it produces electric eld What is a direction of the eld close to the surface The eld must be perpendicular to the surface If there were a parallel component to E charges would experience a force and accelerate along the surface and it would not be in equilibrium The potential energy per unit charge Uqo is the electric potential 3 Electric potential difference between AU 2 W fA F I as O B A The electrical potential difference between A and B equals the work per unit charge necessary to move the charge q from A to B The electric potential difference is independent of the path between points A and B W thVAB Work also is independent of the path Electrostatic forces are conservative The electric potential with respect to infinity work needed to move a unit charge from infinity to point r in the electric field produced by source charge q av Eav E av Electric field EX 2 a XVEZa z ELECTRICAL POTENTIAL IN A CONSTANT FIELD E A B AVABAUABq The electrical potential difference between A and B equals the work per unit charge necessary to move the charge q from A to B AVAB VB VA WAB q I Eds But E constant and End E d then AVABIEods IEdzEfdzEL AVABZEL IAUABqEL I Work to be done to bring the point charge qt from in nity to the point of distance r from the source point charge q W ntG 39 dl ntrdr Z kqtqi W qut 1 Vr Wqt k q r The electric potential due to several point charges is the sum of the potentials due to each individual charge superposition principle The sum is the algebraic sum VkGZ VOatroo VDue to a Charged Conductor Consider two points on the surface of the charged conductor as shown E is always perpendicular to the displacement ds Therefore EOdS 0 Therefore the potential difference between A and B is also zero Vis constant everywhere on the surface of a charged conductor in equilibrium AV 0 between any two points on the surface The surface of any charged conductor in electrostatic equilibrium is an equipotential surface Notice from the spacing of the positive signs that the surface charge density is nonuniform l y 6 Mi An example of a typical problem Point charges q and Q are positioned as shown If q 20 nC Q 20 nC a 30 m and b 40 Hi What is the electric potential difference VA VB quot 39 39 39 quotT 39f 39 900T I I I z I I I r flk39QO 1 Q J39I L o parallel plate capacitor Electric Field gt E lt f 1d 1 cgonSOA 6 Q A 16 scance A l elds Q E add Potential Difference gt d VZE dZQ 180A Q lt elds cancel Wd80A Q22 dQ2280A UdQ2280A12 sOEZAd uE UAd 12 80 E2 true for arbitrary E EQA80 VEdQdA80 CapacitanceCQV CQVsOAd An airfilled capacitor consists of two parallel plates each with an area of 760 cmz separated by a distance of 190 mm A 250V potential difference is applied to these plates a Calculate the electric eld between the plates V E 1 Q b Calculate the surface charge density K E680 80886XIOHCWNm2 1 c Calculate the capacitance O CQVAd d Calculate the charge on each plate GQA Energy Stored in a Capacitor Suppose we have a capacitor with charge q and 0 Then we increase the charge by dq and We must do work dW qu to increase charge Le dW q dq C 0 Integrating q from 0 to Q we can nd the total stored potential electric energy Q Q q 2 U szszjEdq U szgicz cyz 0 0 Capacitors in Parallel Suppose there is a potential C1 q1 difference V between a and b a E j b Then qlVC1 ampq2VC2 C2q2 39 l1 VC1 amp Q2 VC2 V Work like one capacitor C split in two parts I I The charge on C is q q1 q2 a C39 39 q 0 ThenCqVq1q2Vq1VqzVC1C2 This is the equation for capacitors in parallel Increasing the number of capacitors increases the capacitance Capacitors in Series C C2 C a 1qu q W b a II I hill b v1 v2 v Here the total potential difference between a and b is V V1 V2 Also V1 lCl q and V2 lCz 1 The charge on every plate C1 and C2 must be the same in magnitude Then V V1 V2 q C1 q C2 1C1 1C2 q orV1C q gt 1C1C1 1C2 This is the equation for capacitors in series Increasing the number of capacitors decreases the capacitance m lift HF Find the following In the figure use C1 2320 uF and C2 1720 uF Ely IVF the equivalent capacitance of the capaCItors In the 39M figure above 0 is a pair of 02 and 6uF mw Hui In uence of electric eld on ow of electrons An electric eld modi es the trajectories of electrons between collisions When E is nonzero the electrons move almost randomly after each bounce but gradually they drift in the direction opposite to the electric eld Electric current 0 F I 0 awhe dq passes throughO in time dt We de ne the electric current as the movement of charge across a given area per unit time Idqdt SI unit of current 1 Cs 1 Ampere Amp The direction of the current is the direction in which positive charges would move Electrons move opposite to the direction of the current Current density 0 0 gt I 0 k Crosssectional area A If current I ows through a surface A the current density J is de ned as the current per unit area J I A After an electron collides with an ion it will accelerate under anE eldWith a eEm Suppose the average time between collisions is t Then the average velocity is Vd a t e E t m This velocity is called the electron drift velocity Current density 0 7 9 Density of electrons n Number of electrons NnAvdAt vdAt In time At all the electrons in the cylinder move out through the right end Hence the charge per unit time the current is IN eAtneAVdAtAtneAVd The current density is JIAnevdne2tmEsincevdeEtm Ohm s Law 39 We found that J nez39cmE that is that the current J is proportional to the applied electric eld E both are vectors J 039 E Ohm s Law 039 is the Conductivity 039 J E Units are Amz divided by Vm AVm It is useful to turn this around and define the Resistivity as p EJ 16 Units of p are VAm 039 and p are dependent only on the material not on its length or area However consider a metal rod of resistivity p V p area A L E pJ VL P IA V pLA I VIR 0 volts Ohm s Law The macroscopic form V I R is the most commonly used form of Ohm s Law R is the Resistance It depends on the material type and shape R p L A Units ohms 2 As p R A L common units for the resistivity p are Ohmmeters Similarly common units for the conductivity 6 l p are Ohm m391 or Mhom Electrical Power Dissipation In travelling from a to b V energy decrease of dq is dU dq V a dq 1 NOW dq I dt Resistance R Therefore dU I dt V Rate of energy dissipation is dU dt I V This is the dissipated power P Watts or Joules sec Hence PIV PIZR V2R


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