no tagsLecture 004: Continous Charge Distributions and Matter in ElectricFieldsSteveSekula, 29 August 2010 (created 29 August 2010)GoalsDiscuss the transition from finite distributions of charge to continuousdistributions of chargeDiscuss what happens to matter in the presence of electric fieldsContinuous Charge DistributionsWe have been dealing with systems with one or two charges, or treatingmore complex systems (e.g. the heart, on the problem set) like a singlepoint charge. Today, we get tough. Today, we discuss the transitions fromfew charges to lots of charges.Nature is build from elementary charges like electrons and protons.However, a typical number in a volume of air or in a solid is Avagadro'sNumber: ! Good luck adding all of those forces and fieldstogether. We need tools to attack these more realistic situations. We needcalculus.To begin, let's discuss the bulk properties of continuous chargedistributions:Volumes of charge: given a volume (3 dimensions) of charge, we willspeak of the volume charge density, , whose units are charge perunit volume, .Surfaces of charge: if the charge is spread out over a surface, insteadof throughout a volume, we will speak of the surface charge density,, whose units are .General Physics - E&M (PHY 1308) LectureNotesGeneral Physics - E&M (PHY 1308) LectureNotes6:02 0 Â 123Ú C=m 3Û C=m 2General Physics - E&M (PHY 1308) - Lecture Notes file:///home/sekula/Documents/Notebooks/PHY1308...1 of 9 08/29/2010 04:40 PMLine of charge: if the charge can be described as distributed along asingle dimension, we will then speak of linear charge density, ,whose units are .Using these concepts lets us summarize the charge distribution propertiesof an extended, continuous system. Instead of saying a charge is here, atsome coordinates, and over there, at some other coordinates, we speakinstead of densities of charge along a line, surface, or throughout a volume.Cement the concept: examplesImagine that I have a cube, whose sides are each of length . In each ofthe following situations, tell me the appropriate bulk property of thesystem:What is the bulk property of the system if there is a charged objectinside the cube whose strength is ? ANSWER: we should beconcerned about the charge per unit volume, or the volume chargedensity of the system. In that case, it will be .What is the bulk property of the system if the same charge isdistributed over the sides of the cube? ANSWER: we should beconcerned about the charge per unit area, or the surface chargedensity. In that case, we need the surface area of the cube. That's. Thus .What is the bulk property of the system if the same charge isdistributed only along the thin corners at each side of the cube?ANSWER: if the charge is isolated to just the corners where each sidemeets, you can imagine the cube as being made from a wireframewhere each wire holds part of the total charge. Thus, the bulk propertywe are concerned with is the linear charge density. We need to knowthe lengths of these "wires". There are 12 such lines that make up thecube. Thus the total length is , and.Summing the fieldsWe've used the Principle of Superposition to add together individual fieldsfrom a series of charges. Now that we're dealing with lots of charge spreadÕ C=m 1:0m 10C Ú 10C)=(1m) 0C=m = (3= 1310C A 1m ) m = 6 Â (2= 62Û 0C=6m 5=3)C=m = 12= (210C L 2 m 2m = 1 Â 1 = 1Õ 0C=12m 5=6)C=m = 1 = (General Physics - E&M (PHY 1308) - Lecture Notes file:///home/sekula/Documents/Notebooks/PHY1308...2 of 9 08/29/2010 04:40 PMout over lines, surfaces, and volumes, we need a better way to do the sums.Calculus - and specifically the integral - gives us this power.Recall that an integral is just given in the limit of a sum of very smallpieces of a problem. For instance, if we have a bunch of small charges andwe want to find the total field at some point, P. Each infinitesimal unit ofcharge, , is responsible for emitting its own infinitesimal piece of thetotal field, . Thus to obtain the total field we must integrate over theindividual infinitesimal fields:We know how to write the electric field of each infinitesimal point of charge- that's just using Coulomb's Law! We have to do a transformation ofvariable in the integral, from to . That's simply done as follows:Thus:This is the easy part. The hard part is being given a physical problem andbeing asked to solve for the field in terms of the geometry of that problem.Attacking real problems will involve translating the geometry of theproblem into something over which you can then integrate more easily.Let's look at an example.Example: a line of chargeConsider a thin wire that carries a charge. The bulk property of the wirethat we are interested in is the linear charge density, . Imagine that thiswire - perhaps a power line or an electric cable of some other sort - liesalong the x-axis. How do we find the electric field at some point P awaydq dE ~E E ~=Zd~dE ~dq dE q q dq r ~=ïdqdE~!d = rÒddqr2kq^Ód = rÒkr2^Ódqdq=r2k qd^E r ~=Zr2k qd^Õ General Physics - E&M (PHY 1308) - Lecture Notes file:///home/sekula/Documents/Notebooks/PHY1308...3 of 9 08/29/2010 04:40 PMfrom the line?INTERPRET: we have to start somewhere. Let's treat the wire as so longthat we don't have to worry about being close to its ends, where thegeometry changes (and is not very line-like anymore). Thus, let's call thisline infinite in length. Let's define our point P, then, as lying at a point inspace . Thus, it lies a distance from the line along the y-axis.The source charge in the problem is the whole wire.DEVELOP: Make a drawing of the problem. We divide the wire into smallcharge elements . NOTE: one thing we see right away is that if weconsider two charge elements lying an equal distance along the x-axisfrom , the electric fields of these two elements is equal and oppositealone the x-axis and cancels. This leaves only the y-components of theindividual fieldsm , to be considered. This immediately reduces theproblem into one where we only have to find the y-component of eachinfinitesimal field! Thus we only need the y-component of each unit vector,.Let's explore that last statement. The distance from any point on the wireto the point P is , and the vector from that point on the wire toP is given by , where is the location along of the charge element, and is the distance from the wire to P along the y-axis. How does oneobtain the unit vector from this? Recall that generally
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