# Lec5 - Physics 8B Professor Catherine Bordel Lecture 5 ASUC...

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Physics 8B Professor Catherine Bordel 09/08/10 Lecture 5 ASUC Lecture Notes Online is the only authorized note-taking service at UC Berkeley. Do not share, copy, or illegally distribute (electronically or otherwise) these notes. Our student-run program depends on your individual subscription for its continued existence. These notes are copyrighted by the University of California and are for your personal use only. D O N O T C O P Y Sharing or copying these notes is illegal and could end note taking for this course. LECTURE Gauss’s Law – the electric flux, Φ E , through a closed surface (s) is equal to the total amount of electric charge Q enc enclosed in the closed surface (s) divided by the permittivity of vacuum ε 0 . = = ) ( 0 s enc E Q A d E ε φ v v The circle through the integral symbol means that the integral is taken over a closed surface. This surface is a figurative surface, meaning it does not exist and does not carry electric charge. Gauss’s Law is one of the four fundamental laws of electromagnetism. I’d like to cover an example right now with you. We’re going to consider a hollow sphere with a charged surface. The center is O, and the radius is R. You have a charge distribution on the surface that is σ , and we assume σ is positive. If it’s negative, then E will point in the opposite direction, but it has the same symmetry and doesn’t change the way you need to solve the problem. We want to calculate the electric field at point M, which is a distance r away from the center O. You see that the physical surface that carries the electric charge is the natural boundary between two regions of space, the region that is enclosed by the sphere and the region outside. We will have to study distances of r>R and also r<R. In each case we need to calculate the flux going through the closed surface that we choose, and we will be able to determine the electric field. The first step you have to do to use Gauss’s Law is to figure out by the symmetry of the charge distribution what the direction of the electric field is going to be. What is going to be the direction of the electric field from this spherical charge distribution?

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## This note was uploaded on 10/02/2010 for the course PHYSICS 8B taught by Professor Shapiro during the Spring '07 term at Berkeley.

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Lec5 - Physics 8B Professor Catherine Bordel Lecture 5 ASUC...

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