Monday, March 22, 2010

Irodov Problem 3.19























The electric flux through the circle is given by,



Where E is the electric field and dS is the area vector through which E passes. From problem 3.14 we already know that E is given by,



Consider an infinitesimally thin ring section of radius r and width dr. The area vector of this infinitesimally thin ring section is given by,



Hence the flux is given by,