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,
Subscribe to:
Posts (Atom)