lect04 - Physics 227: Lecture 4 Applications of Gausss Law,...

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Unformatted text preview: Physics 227: Lecture 4 Applications of Gausss Law, Conductors in Electrostatics Lecture 3 review: Calculate the electric feld through superposition as a sum oF charges or integral over a charge distribution. Symmetry arguments can give you the direction oF the feld, in some cases. Motion is calculated From the (electric) Force, as you learned beFore. lux: = E . dA = E dA = E cos dA. Gausss Law: For a closed surFace, = q enclosed / . The ux does not depend on the size or shape oF the Gaussian surFace, or the position oF the charges, it depends only on the total enclosed charge. Thursday, September 15, 2011 Applying Gausss Law Choose a surface that reFects the symmetry of the charge distribution. You want a surface for which E is some constant or 0! or E 0, calculate the area of the surface A. Then = E dA = E A. But = q enclosed / . So... E = q enclosed /A . Thursday, September 15, 2011 Gausss Law for Point Charge q Use a spherical Gaussian surface centered on the point charge q. Why? The electric Feld at the surface will be constant and perpendicular to it. Gausss Law: = E . dA = q/ E . dA = E dA = E . A = 4 r 2 E 4 r 2 E = q/ E = q/4 r 2 This is what we learned from Coulombs Law + the deFnition of the electric Feld: E = C /q test = (qq test /4 r 2 )/q test = q/4 r 2 Thursday, September 15, 2011 Gausss Law for Uniform Sphere of Charge Use a Gaussian sphere co-centered with the uniform sphere of charge Why? The electric Feld at the surface will be constant and perpendicular to it. Gausss Law: = q/ = (4/3) r 3 / . E . dA = E dA = E . A = 4 r 2 E. E = r/3 . Does this agree at the surface with E = q/4 r 2 ? E = r /3 = qr/((4 r 3 /3)(3 )) = qr /(4 r 3 ) q/4 r 2 . Yes, it does. Outside the sphere, r > r , the Feld is identical to that of a point charge. But inside the sphere......
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lect04 - Physics 227: Lecture 4 Applications of Gausss Law,...

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