Is the Electric Field Normal to the Surface?

[auk411]

Homework Statement

Let's say you have a solid cylinder. It is nonconducting. It is long enough such that the contributions from the ends are negligible. It has a non-uniform volume charge density of Ar^2, where A is some positive constant. Then you are asked to find the electric field at different radii, respectively.

This is a problem that I was given as homework. What confuses me is that the authors assume that the Flux, F, = E(Surface Area of curved area). That is, F = E2\piRH, where H is the height of the cylinder. Typically, this I have no problems with this; I understand what is going on (typically). However, in this case I see nothing in the problem that warrants the assumption that the flux can be rewritten as F=Ecos(0)2\piRH. Am I missing something here? Couldn't the Electric field be written as <x,y,z>. In which case, the the angle between the Electric field and the normal vector pointing out from the surface would NOT be 0. Thus, making the assumption false.

Is it the case that anytime you have a gaussian surface that is a cylinder, the Electric field will always be parallel to any normal vector to the surface. Or is there something in the problem that warrants the assumption that this is true?

Btw, the problem only contains the following information: that is a cylinder, it is really long, it has a non-uniform charge density = Ar^2. A = 2.5\mu C/m^5, the radius of the cylinder is .04 m. And we are to find the electric field at r = 3 cm and r = 5 cm.

Given that I make the assumption that the angle between E (vector, not magnitude) and a normal vector to the (curved) surface is 0, I can solve the problem. However, I can't find any reason to make the assumption.

Any explanation would be great.

Homework Equations

The Attempt at a Solution

Homework Statement

Homework Equations

The Attempt at a Solution

 

[ehild]

You are right, nothing ensures that the electric field is normal to the surface of the cylinder, but symmetry. If the cylinder had finite length, it would not be true. If the charge distribution depended on height or the azimuthal angle, it would not be right.
If it was a metal surface, the electric field would be normal to the surface as the electric field is the negative gradient of the potential, and a metal surface is a equipotential surface (in the static case). This cylinder is not from metal, but both the geometry and the charge distribution has the cylindrical symmetry. With this assumption, you can find a solution which fulfils the Maxwell equations, and as the solutions of these equations are unique, this is the solution.

ehild