What does the electric field caused by a charged cylinder look like? I am talking about a cylinder with a large enough radius:length ration that it cannot just be treated as a line. I would think that the electric field lines come out the curved and flat surfaces orthogonally, and they remain straight. Is this correct?
 A: Let's analyse the case for an infinitely long cylinder first, then move to a finite one (I've just realised you mean one of finite length having written my answer). Your hypothesis is right only in the case of an infinitely long cylinder, but not for one of finite length, unless your cylinder is a perfect conductor, in which case no symmetry assumptions are needed to make your hypothesis true.
Infinitely Long Cylinder Case
In this case you are right, by dint of the following symmetries. You must assume, however, (but you're likely aware of this) that the charge distribution on the cylinder is a function of the radial co-ordinate $r$ only; here $r$ is the distance of the point in question from the cylinder's axis of rotational symmetry. Then you must argue using the following four symmetries begotten by the uniformity of charge:


*

*The system's translational invariance: with space labelled by cylidrical polar co-ordinates $(r,\,\theta,\,z)$ with the $z$-axis along the cylinder's axis, the electric field distribution $\vec{E}(r,\,\theta,\,z)$ must fulfill $\vec{E}(r,\,\theta,\,z+u)=\vec{E}(r,\,\theta,\,z),\,\forall\,u\in\mathbb{R}$, whence $\vec{E}(r,\,\theta,\,z)=\vec{E}(r,\,\theta,\,0)$ and so $\vec{E}$ is not a function of $z$;

*The system's rotational symmetry about the cylinder's axis. This means that the pattern of field lines must be invariant with respect to a rotation through any angle about the axis. Let $\vec{E}(r,\theta)$ define the electric field distribution. Then our symmetry means that $\vec{E}(r,\,\theta + \phi) = \mathbf{R}(\phi)\,\vec{E}(r,\,\theta)$ where $\mathbf{R}(\phi)$ is $2\times2$ rotation matrix for a rotation through angle $\phi$ about the $z$-axis. Therefore $\vec{E}(r,\,\theta) = \mathbf{R}(\theta)\,\vec{E}(r,\,0)$ and so $\vec{E}$ wholly is defined by its values $\vec{E}(r,\,0)$ along the ray $\theta = 0$.

*So it remains to show that $\vec{E}(r,\,0)$ is indeed directed along the radial ray $\theta = 0$. To show this, we need to call on the system's reflexional symmetry (i) about the plane $\theta = 0$ and (ii) about the plane $z=0$. $\vec{E}(r,\,0)$ can only be invariant with respect to a reflexion about $\theta = 0$ if it has no component in the $\hat{\theta}$ direction. Likewise it can only be invariant with respect to a reflexion about $z=0$ if it has no $\hat{\mathbf{z}}$ component. So we have, at long last, proven that:
$$\vec{E}(r,\,\theta,\,z) = E(r,0,0) \hat{\mathbf{r}}$$
and we're all done!
Finite Length Cylinder
So now we repeat the above, but now we have lost (i) our translational invariance in (1) above, so that our field can now be a function of $z$ and (ii) our reflexional symmetry about any plane of constant $z$, aside from the special case of $z=0$ if we make $z=0$ the plane halfway along the cylinder's length. So, on working through the above argument without these two lost symmetries we get:
$$\vec{E}(r,\,\theta,\,z) = E_r(r,0,z) \hat{\mathbf{r}} + E_z(r,0,z) \hat{\mathbf{z}}$$
i.e. there is no $\hat{\theta}$ component but the field on each plane of constant $z$ is still defined wholly by the field along the $\theta=0$ ray. We have one last symmetry: the relfexional symmetry about $z=0$ (unlike for other planes of constant $z$). On requiring inavriance with respect to this symmetry, we get $E_z(r,0,z) = -  E_z(r,0,-z)$ and $E_r(r,0,z) = E_r(r,0,-z)$, i.e. the radial and axial components are even and odd functions of $z$, respectively. So here, unless your cylinder is a perfect conductor, the field will not in general be orthogonal to the cylinder's surface.
