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Unformatted text preview: Physics 122 Spring 2009 – Document #12: Cycle 2 Review Sheet page 1 of 10 PHYS 122: Cycle 2 Review Sheet February 8, 2009 Cycle 1 Materials Physics is a cumulative subject and the this is especially so for the Cyclic approach. Therefore is it explicitly the case that each student are responsible for all materials delineated in the Cycle 1 Review Sheet (Handout Document #05) . As part of your preparations for the Second Hour Exam be sure you are completely comfortable with all materials presented in this previous review sheet as any question on any of these topics is “fair game” for the next exam. Coulomb’s Law and The Electric Field for Continuous Charges We reiterate a major central fundamental concept for the course: any collection of charge creates an associated electric field . We can calculate explicitly and exactly the value of the electric field at any point by applying an extended form of Coulomb’s Law which indicates that the total electric field is the field that results when one integrates over all of the contributions of each little bit of charge in a continuous charge distribution. The contribution to one bit of charge dq is given by: d vector E = 1 4 πǫ dq r 2 ˆ r and so vector E = integraldisplay allspace d vector E = 1 4 πǫ integraldisplay allspace dq r 2 ˆ r You should understand the basic technique of determining the electric field by evaluating this integral. This integral, which in general needs to be done over three spatial dimensions ( dq = ρ ( vector r ) dxdydz ) for each of three vector components in the field can be technically complex, even for relatively simple charge distributions, although using symmetry can help. Electric Flux: In order to apply Gauss’ Law we need to introduce a new physical quantity: the Electric Flux. In class I showed I this quantity is analogous to the volume flow rate of water through the opening of a pipe. There are two mathematically equivalent concept for the electric flux: • In the special case that the electric flux is constant in magnitude and has a fixed orientation angle with respect to some surface, then the flux is equal to the dot product of the vector electric field and the area vector associated with any surface. We define the area vector vector A as a vector that points normal (e.g. perpendicular) to the surface and has magnitude equal to the area of the surface. In this case the electric flux is defined: Φ E = vector E · vector A = EA cos θ Where θ is the angle between vector E and vector A . In the case that vector E and vector A are aligned, Φ E = EA . In the case that vector E is perpendicular to vector A , Φ E = 0 . Physics 122 Spring 2009 – Document #12: Cycle 2 Review Sheet page 2 of 10 • If the surface is curved, the electric field nonconstant, or there is any other complication, we can still define the total electric flux as the integral over the surface for each element of area dA as follows: Φ E = integraldisplay surface vector E · vector...
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This note was uploaded on 11/06/2009 for the course PHYS 122 taught by Professor Raman during the Spring '09 term at Lehigh University .
 Spring '09
 Raman
 Physics

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