# SMU PHYS 1308 - Continous Charge Distributions and Matter in Electric Fields (9 pages)

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**View the full content.**## Continous Charge Distributions and Matter in Electric Fields

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## Continous Charge Distributions and Matter in Electric Fields

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- Pages:
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- School:
- Southern Methodist University
- Course:
- Phys 1308 - General Physics Ii

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General Physics E M PHY 1308 Lecture Notes le home sekula Documents Notebooks PHY1308 General Physics E M PHY 1308 Lecture Notes Lecture 004 Continous Charge Distributions and Matter in Electric Fields SteveSekula 29 August 2010 created 29 August 2010 no tags Goals Discuss the transition from finite distributions of charge to continuous distributions of charge Discuss what happens to matter in the presence of electric fields Continuous Charge Distributions We have been dealing with systems with one or two charges or treating more complex systems e g the heart on the problem set like a single point charge Today we get tough Today we discuss the transitions from few 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 s Number 6 02 1023 Good luck adding all of those forces and fields together We need tools to attack these more realistic situations We need calculus To begin let s discuss the bulk properties of continuous charge distributions Volumes of charge given a volume 3 dimensions of charge we will speak of the volume charge density whose units are charge per unit volume C m3 Surfaces of charge if the charge is spread out over a surface instead of throughout a volume we will speak of the surface charge density whose units are C m2 1 of 9 08 29 2010 04 40 PM General Physics E M PHY 1308 Lecture Notes le home sekula Documents Notebooks PHY1308 Line of charge if the charge can be described as distributed along a single dimension we will then speak of linear charge density whose units are C m Using these concepts lets us summarize the charge distribution properties of an extended continuous system Instead of saying a charge is here at some coordinates and over there at some other coordinates we speak instead of densities of charge along a line surface or throughout a volume Cement the concept examples Imagine that I have a cube whose sides are each of length 1 0m In each of the following situations tell me the appropriate bulk property of the system What is the bulk property of the system if there is a charged object inside the cube whose strength is 10C ANSWER we should be concerned about the charge per unit volume or the volume charge density of the system In that case it will be 10C 1m 3 10C m3 What is the bulk property of the system if the same 10C charge is distributed over the sides of the cube ANSWER we should be concerned about the charge per unit area or the surface charge density In that case we need the surface area of the cube That s A 6 1m2 6m2 Thus 10C 6m2 5 3 C m2 What is the bulk property of the system if the same 10C charge is distributed only along the thin corners at each side of the cube ANSWER if the charge is isolated to just the corners where each side meets you can imagine the cube as being made from a wireframe where each wire holds part of the total charge Thus the bulk property we are concerned with is the linear charge density We need to know the lengths of these wires There are 12 such lines that make up the cube Thus the total length is L 12 1m 12m and 10C 12m 5 6 C m Summing the fields We ve used the Principle of Superposition to add together individual fields from a series of charges Now that we re dealing with lots of charge spread 2 of 9 08 29 2010 04 40 PM General Physics E M PHY 1308 Lecture Notes le home sekula Documents Notebooks PHY1308 out 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 small pieces of a problem For instance if we have a bunch of small charges and we want to find the total field at some point P Each infinitesimal unit of charge dq is responsible for emitting its own infinitesimal piece of the Thus to obtain the total field we must integrate over the total field dE individual infinitesimal fields E Z dE 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 of to dq That s simply done as follows variable in the integral from dE dE dE dq dq k dq k dq d kq r r dq dq 2 r 2 2 dq dq r r r Thus E Z k dq r r2 This is the easy part The hard part is being given a physical problem and being asked to solve for the field in terms of the geometry of that problem Attacking real problems will involve translating the geometry of the problem into something over which you can then integrate more easily Let s look at an example Example a line of charge Consider a thin wire that carries a charge The bulk property of the wire that we are interested in is the linear charge density Imagine that this wire perhaps a power line or an electric cable of some other sort lies along the x axis How do we find the electric field at some point P away 3 of 9 08 29 2010 04 40 PM General Physics E M PHY 1308 Lecture Notes le home sekula Documents Notebooks PHY1308 from the line INTERPRET we have to start somewhere Let s treat the wire as so long that we don t have to worry about being close to its ends where the geometry changes and is not very line like anymore Thus let s call this line infinite in length Let s define our point P then as lying at a point in space x y 0 y Thus it lies a distance y 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 small charge elements dq NOTE one thing we see right away is that if we consider two charge elements lying an equal distance x along the x axis from x 0 the electric fields of these two elements is equal and opposite alone the x axis and cancels This leaves only the y components of the to be considered This immediately reduces the individual fieldsm dE problem into one where we only have to find the y component of each infinitesimal field Thus we only need the y component of each unit vector r y y r Let s explore that last p statement The distance from any point on the wire to the point P is r x2 y 2 and the vector from that point on the wire to P is given by r x y where x is …

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