Water molecule $\text{H-O-H}$ angle in electrostatic field An image showing water molecule within a cylinder with positive charge on the inside and negative on the outside. To what extent will the $\text{H-O-H}$ angle adjust itself to the electrostatic force? Scaling this up to that the cylinder is not just a few hundred picometers, but a few micrometers, 10^5 times, is there still an effect?

 A: Here is an answer, for long enough distances/small fields the water molecule is neutral, better modeled as a dipole but does not really influence the angle. For strong fields there might be an influence and that it is worth investigating. The answer is complicated as it will depend on experimental feedback, quantum mechanics, contact forces and, most importantly, solving the problem for different geometries.
However in your image above you have chosen a very symmetric case. The axial symmetry of the problem allows to calculate the electric field very easily. Mike S already pointed to the answer: Electric Field of Hollow Cylinder . I will limit myself to provide you the main result that concerns you.
Let us say that the angle of water molecule when the cylinder is not charged is $\theta_0$. When is charged it becomes $\theta_0+\Delta\theta(d)$, where $d$ is the distance. When you solve for the electric field in the cylinder, you will find that the $\Delta\theta(1\;\mathrm{pm})=\Delta\theta(1\;\mu\mathrm{m})$ (strict equality).
In conclusion, for the geometry given, the angle does not change with distance, the rest is up to you to see why.
