Recent content by annoyingdude666

hello SammyS thank you for answering my question btw, i still don't understand. what do you mean by "more convenient" or "more simpler" ? can u explain why ? because from what i know, the charges of conducting rod spread on its surface (surface = area), so why we use lambda rather than sigma ...

hi, i still don't understand why infinite thin-walled cylindrical shell or conducting rod use lambda rather than sigma ? lambda = C/m ,,, sigma = C/m^2 i mean when we look at conducting rod, the charges inside the conductor is zero, so the charges spread on the surface of conducting rod(have...