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## Main Question or Discussion Point

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 same form as thin-walled cylindrical shell), which is when we calculate the electricfield use gauss's Law using formula :

integral(E . dA) = q enclosed / vaccum permitivity

the q enclosed is sigma times the surface area , rather than lambda times length ?

i find this from "Fundamental of Physics" Halliday Resnick

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 same form as thin-walled cylindrical shell), which is when we calculate the electricfield use gauss's Law using formula :

integral(E . dA) = q enclosed / vaccum permitivity

the q enclosed is sigma times the surface area , rather than lambda times length ?

i find this from "Fundamental of Physics" Halliday Resnick