An infinite cylinder of radius \"a\" and aligned with the z-axis is filled with
ID: 2144756 • Letter: A
Question
An infinite cylinder of radius "a" and aligned with the z-axis is filled with a charged gas. The volume charge density inside the cylinder is:
Pv = 1 - e^( sqrt( x^2 + y^2 ) - a )
Derive expression for E and V inside the cylinder. Assume that the potential V at the surface of the cylinder is 0.
I understand that since it is an infinite cylinder, the only electric field experienced will be in the unit vector "r" direction due to symmetry. I'm just having a hard time figuring everything else out.
Explanation / Answer
dv = 2 pi r' L dr'
dq = Pv dv = (1 - e^(sqrt(x^2 + y^2) - a)) (2 pi r' L dr')
sqrt(x^2 + y^2) = r
==> dq = Pv dv = (1 - e^(r' - a)) (2 pi r' L dr')
integrate to find q:
q = int{Pv dv} ; from r'=0 to r'=r
==> q = int{(1 - e^(r' - a)) (2 pi r' L dr')} ; from r'=0 to r'=r
==> q = (2 pi L) (r'^2/2) - (2 pi L) (r' - 1) e^(r' - a) ; from r'=0 to r'=r
==> q = {(2 pi L) (r^2/2) - (2 pi L) (r - 1) e^(r - a)} - {(2 pi L) (0^2/2) - (2 pi L) (0 - 1) e^(0 - a)}
==> q = (2 pi L) (r^2/2) - (2 pi L) (r - 1) e^(r - a) - (2 pi L) e^(a)
==> q = (2 pi L) ((r^2/2) - (r - 1) e^(r - a) - e^(a))
Gauss's law:
int{E.da} = q/(epsilon0)
==> E (2 pi r L) = ((2 pi L) ((r^2/2) - (r - 1) e^(r - a) - e^(a)))/(epsilon0)
==> E = (r^2/2 - (r - 1) e^(r - a) - e^(a))/(r epsilon0)