PHYS 110A Lecture Notes - Lecture 2: Divergence Theorem, Volume Integral, Point Particle
Document Summary
Check solution to line charge problem: units: :c/m, 0:c^2/nm^2 -> e:n/c which checks out, sign is correct, limits: when d>>l, e=q/4 0d^2. It acts as a point charge at far distances, which is correct. As d approaches 0, e = /4 0d which diverges as d, not d^2. Define coordinate system where z axis lies the line between point p and the sphere"s center. Therefore e_x and e_y are both equal to zero due to symmetry. With substitution of charge per volume, we get: E = ( /4 0) integral of (r^2 sin d d dr (z-rcos )/(z^2+r^2-2zrcos )^(3/2)) Most common notation conveys field strength by density of the lines. Flux: rate of flow of ___ across a given area. _e = integral of (edacos ) = (q/4 0) integral of (cos da/r^2) = q/ 0. _e = integral of sum of (e_i*dacos ) = sum of q_i/ 0 = q_enclosed/ 0.
