Chemistry - Why is H₂O V shaped?
The oxygen atom fills its octet rule by forming two bonds. It shares 1 electron in a covalent bond with each hydrogen and has 4 remaining valence electrons. It is sp$^3$ hybridized and has the 4 non-bonding electrons in two lone pairs. An sp$^3$ hybridized atom has four attachment points spaced approximately 109$^\circ$ apart and has the shape of a pyramid with a triangular base. See this image for an example. You don't typically see the water molecule drawn in 3 dimensional space with the lone pairs of electrons, all you see is a planar molecule with H-O-H bond angle of ~109$^\circ$.
It is not the + charges on hydrogens, but the negative charges on the O-H bonds, and on the O it self that is dominant in determining the molecule's shape. There are four non-bonding electrons on O -two lone pairs- on top of the two O-H bonds.
So you should think of four groups around O, not two.
I mean, if the hydrogens have a partial positive charge, then they should try to get away from each other.
Well, this is not a complete answer to your question, but this could be a factor I guess.
Bond pair – lone pair repulsion is always greater than bond pair – bond pair repulsion
So what I want to say is that the repulsion between the O---H bond and the lone pair is so heavy such that it is really hard for two hydrogen atoms to get away each other as you are expecting to happen.