Want to use a battery?

Firstly, one must understand how can a battery lead to rise in height of water.

A coaxial metal cylinder will act as a cylindrical capacitance with a potential difference between the two cylinders. We have a formula for that -

$C_0 = \frac{2\pi \epsilon_{0}h}{\log\left(\frac{b}{a}\right)}$

(I am not deriving it here, if you want, just leave a comment below).

Now when I connect a battery of, let's say, $V$ volts, the energy stored in the capacitor in the form of electrostatic field is

$U = \frac{{(CV)}^2}{2C}$

If a dielectric of relative permittivity $\epsilon_r$ is put between the separation of the two cylinders upto a height $x$ , then the capacitance is given as a parallel combination of two capacitances -

$C_1 = \frac{2\pi \epsilon_{0}(h-x)}{\log\left(\frac{b}{a}\right)}$ , $C_2 = \frac{2\pi \epsilon_r \epsilon_{0}x}{\log\left(\frac{b}{a}\right)}$

Therefore, $C = \frac{2\pi \epsilon_{0}(h + (\epsilon_r -1)x)}{\log\left(\frac{b}{a}\right)}$

When this capacitor is brought close to the water surface(just touching it), the water surface feels an inward force which is due to the loss in potential energy of the capacitor if the dielectric(water) is brought into it. Neglecting edge effects, this force is equal to

$F = - \frac{dU}{dx} = -\frac{d}{dx} \left(\frac{Q^2}{2C}\right)$

since charge on the capacitor remains conserved.

$= -\frac{{(C_0 V)}^2}{2} \frac{d}{dx} \left(\frac{\log\left(\frac{b}{a}\right)}{2\pi \epsilon_{0}(h + (\epsilon_r -1)x)}\right)$

$= \frac{V^2 \pi \epsilon_0(\epsilon_r -1)}{\log\left(\frac{b}{a}\right) \left(1 + (\epsilon_r -1)\frac{x}{h}\right)^2}$

If this upward force has given rise to water upto height $x$ , then this must be balancing the risen water's weight.(obviously we have assumed that angle of contact of water-metal interface is equal to $90^{\circ}$ .

$\rho \pi(b^2-a^2)g x = \frac{V^2 \pi \epsilon_0(\epsilon_r -1)}{\log\left(\frac{b}{a}\right) \left(1 + (\epsilon_r -1)\frac{x}{h}\right)^2}$

Now, this rise( $x$ ) should be equivalent to that of a glass coaxial cylinder.

For the same, force of surface tension acts perpendicularly upwards(angle of contact = $0^{\circ}$ for glass-water interface) along the inner and outer cylinder surfaces.

Using FBD of water, this must balance the weight of the risen water.

$\rho \pi(b^2-a^2)g x = 2 \pi (b+a) \sigma$

$x = \frac{2 \sigma}{(b-a)\rho g}$

Using this in our previous formula for $V$ ,

$V^2 = \left(1 + (\epsilon_r -1)\frac{2\sigma}{(b-a)\rho g h}\right)^2 \frac{2(b+a)\sigma \log\left(\frac{b}{a}\right)}{\epsilon_0 (\epsilon_r -1)}$

Now, just take a calculator and put in the values, we get $V \approx 3958.57 V$

For our purposes, that becomes $4 kV$ .