This is a homework question, so please don't give me the answer outright. I just need help conceptually.
"A cylindrical shell of length 190 m and radius 4 cm carries a uniform surface charge density of σ = 12 nC/m2.
(a) What is the total charge on the shell?
Find the electric field at the ends of the following radial distances from the long axis of the cylinder.
(b) r = 2 cm"
To part (a) I answered 573nC
circumference*length*(charge density)=(2pi*.04)*190*12E-9=573E-9
Part b is obviously very difficult to calculate absolutely correctly, since there is no obvious gaussian surface perfectly orthogonal to the electric field at every point. However, it seems to me that a short cylinder with its axis aligned with the charged cylindrical surface would make a good approximation, and I believe that is what is expected of me in solving this problem.
So I'll do it with a cylinder of length 1m for simplicity's sake. Since (to a good approximation) the field will be orthogonal to the cylinder's curved surface at all points, and the flux through the ends will be 0, the flux through the curved survace will equal the field strength times the surface area of this surface, which will also equal the charge enclosed over epsilon naught (natural permittivity of space, I'll write this e0).
E*A=Q/e0
E=Q/(e0*A)
So the charge enclosed will simply be the charge in a 1m section of the charged cylinder.
Q=(2pi*.04)*12E-9
The surface area of my Gaussian surface is…
A=(2pi*.02)
(ends not included because the flux through the ends of the cylinder is 0)
E=.04*12E-9/(en*.02)=2710.581760219553
When I put this answer into Webassign (online homework, graded immediately) it is marked as wrong. My answer to part (a) is, however, correct. What is my problem?
Best Answer
It's a stupid mistake. When r is less than the radius of the cylinder, the electric field will be 0. I solved the problem not taking this into account and just plugged in the numbers. My method is perfectly valid for any r greater than 4cm.