Water has a very high dc dielectric constant (relative permittivity ε = 80), which should lead to some interesting characteristics in an electric field.
The stored energy in a parallel plate capacitor is W = ½CV2 = ½εε0AV2/d
where ε0 is the permittivity of free space, A = plate area, and d the separation.
If we have a parallel plate capacitor roughly half submerged in ultrapure water, such that the electric field is horizontal along x, the plate width is y, separation d (along x), and the plate height is h. Plate area A = yh. If the depth of the capacitor plate in the water is z, then the total capacitance is
C = ε0y[εz + (h-z)]/d
When the voltage is turned on, the vertical force pulling the water up between the plates is
Fz = ∂W/∂z = ½ε0y[ε-1]V2/d
This upward force is balanced by the downward gravitational force Fg on the water raised up a distance z-z0 between the plates, where z0 is the equilibrium height of the water in the capacitor gap with the voltage off. Fg = yd(z-z0)ρg, where ρ is the density of water.
Setting Fz = Fg and rearranging, we get (note: y cancels out, so y= ~ 5 cm is adequate):
z-z0 = ½ε0(ε-1)V2/d2ρg
Using ε0= 8.85 x 10-12 Farads/m, d=0.003 m, V = 1000 volts, ε=80, ρ=1000 Kg/m3, g = 9.81 m/s2,
we get z-z0 = 0.004 m (=4 mm)
The separation d is sufficient to permit the water level in gap between the plates reach equilibrium (Adding some photographic Photo Flo to the water might be useful), and the electric field is about 3300 volts per cm (not enough to spark). Increasing V by a factor of 2 will increase z-z0 by a factor of 4.
So an electric field has an effect on water.
Bob S