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Unformatted text preview: 1 Physics 9HD Midterm 1 Review, Fall 2006 Note: All units in CGS, unless otherwise specified. e = charge of electron = 4.8 x 10 -10 esu = 1.6 x 10 -19 coul Electric Forces and Fields, Gausss Law Coulombs Law : r r Q Q F 2 2 1 = r , i Q in esu, r in cm, F in dynes Electric Field (in dyne/esu = esu/cm 2 ): 1 2 1 1 r r Q E = r for point charge, E q F r r = Electric Field between 2 parallel plates : 4 = E r , where =surface charge density = Q/A Superposition of electric fields: n Tot E E E E E r r r r r + + + + = ... 3 2 1 E r field of a charge distribution: + + = = 2 2 2 2 ) ( ) ( ) ( ) , , ( ) , , ( ) , , ( z z y y x x z d y d x d r z y x r z d y d x d r z y x z y x E r , where r is a unit vector pointing from ) , , ( z y x to ) , , ( z y x . Gausss Law: = = = = dl da dv Q a d E encl 4 4 4 4 r r Differential form: 4 = E r r If you are using Gausss Law to find electric field, you must choose surface so that E r is constant on the surface. (Conversely, if the problem is not symmetric enough that you can identify a surface over which you expect E r to be constant, Gauss s Law will not help you to find the...
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