This preview shows page 1. Sign up to view the full content.
Unformatted text preview: the endcaps of the cylinder. Plugging all of this into Gausss law gives: E = q enc H dA = A 2 A = 2 Notice that this diers by a factor of 2 from the equation for the electric eld at the surface of a conductor. The factor of 2 comes in because inside of a conductor, the electric eld is zero and hence the total area which contributes to the ux is only A instead of 2 A . Otherwise, the derivation is the same. 4.4.3 A Uniformly Charged Sphere: Spherical Symmetry In this case we choose a spherical Gaussian surface centered at the center of the sphere to mimic the symmetry. Lets rst consider the eld outside of the sphere. Because of symmetry, the eld must point radially outward and must be constant at equal distances from the sphere. The total 9...
View Full
Document
 Spring '08
 Any
 Physics, Charge

Click to edit the document details