You can use [itex] EI \frac{d^{3} \nu}{dx^{3}} = V [/itex] as long as the cross section of the beam is constant. Where [itex] \nu [/itex] is the deflection.
Rectangular cross-section. The crossection will only affect I, in the formula. We can just keep it as I, so we don't have to worry about the cross-section.
The shear stress would induce a tensile stress in the beam with maximum tensile stress at the fixed end and only shear at the free end. Since the shear load is only on the upper surface, it is unbalanced, so it would induce a bending moment in the beam.
To make sure I understand what you have; a constant rectangular cross section with a uniform horizontal force applied only along the top surface of the section and perpendiular to the longitudinal axis of the beam, and the beam has one end fixed and one end free.
Your deflection can be broken into components of that induced by horizontal shear, bending moment and torsional moment. The shear and bending moment would give translational displacements and the torsional moment generates rotational displacement about the long. axis of the beam. Usually, the bending moment will give the significant deflection of a real world beam and the shear and torsional deflection are often minimal in comparison and sometimes ignored. But you certainly cannot ignore the stresses which they produce.
The stresses will be from the direct shear, torsional shearing, and tension and compression from the bending moment. Your highest stresses will be at the fixed end.