## Electric field intensity inside parallel-plate capacitor filled with water

First, I’d like to say that this is not homework, but my research work. I’ll be very appreciated if someone could help me to address this issue, many thanks in advance.

My experimental purpose is to generate uniform electric field inside microchannel by externally-applied electric field. (“externally” here means no direct contact between electrodes and water)

But there is a questionable point still killing me. As I know, in pure water, autoionization of water molecules can generate hydroxide and hydronium ions at a constant concentration, and these free ions can form Debye space charge layers that screen electric fields. Does it mean, no matter how high DC V is (like 1 KV), the final electric field intensity inside microchannel would go to zero after instantaneous screening process?

If the above statement is valid, then I apply alternating current voltage, like sinusoidal AC (100 kHz, peak voltages from 0 and 500 V) instead of DC, I wonder how to calculate electric field intensity inside microchannel? Taken the rate of electrostatic screening into account, what the minimal frequency of ac should I set ?

Many thanks again for your attention to this matter.
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 If you know what current needs to flow you can calculate the frequency you need. Read up on impedance http://en.wikipedia.org/wiki/Electrical_impedance Basically if r is the resistance and c the capacitance of your setup and f is the frequency then the impedance z is z = sqrt(r^2 + (1/(2*pi*f*c))^2) and i = u/z So with that you can then calculate what voltage and frequency you need to sent a certain current through your capacitor.

 Quote by DrZoidberg If you know what current needs to flow you can calculate the frequency you need. Read up on impedance http://en.wikipedia.org/wiki/Electrical_impedance Basically if r is the resistance and c the capacitance of your setup and f is the frequency then the impedance z is z = sqrt(r^2 + (1/(2*pi*f*c))^2) and i = u/z So with that you can then calculate what voltage and frequency you need to sent a certain current through your capacitor.

Do you think, when DC is applied, the induced electric field by H+ and OH- ions inside H2O-filled microchannel would be instantaneously strong enough to balance the external electric field? Then no net electric field inside microchannel?

Mentor

## Electric field intensity inside parallel-plate capacitor filled with water

I think you need numbers for the mobility of OH- and H3O+ ions in water, driven by an electric field. If the time-scale there is longer than the time-scale for diffusion*, you are probably fine, otherwise the electric field strength will decrease over time and you need AC which is quicker than that.
Keep in mind that the main voltage drop will occur in the dielectric plates anyway.

*I played a bit with equations, and think that the equilibrium is an exponential function for the concentration, with a typical length given by the ratio of "influence of the electric field" to diffusion. If this typical length is >>400µm, DC is fine. If it is <<400µm, DC would give nearly no field inside, apart from the boundary regions, between those regions it is something in between.

 Quote by mfb I think you need numbers for the mobility of OH- and H3O+ ions in water, driven by an electric field. If the time-scale there is longer than the time-scale for diffusion*, you are probably fine, otherwise the electric field strength will decrease over time and you need AC which is quicker than that. Keep in mind that the main voltage drop will occur in the dielectric plates anyway. *I played a bit with equations, and think that the equilibrium is an exponential function for the concentration, with a typical length given by the ratio of "influence of the electric field" to diffusion. If this typical length is >>400µm, DC is fine. If it is <<400µm, DC would give nearly no field inside, apart from the boundary regions, between those regions it is something in between.

I once used the following equations to calculate Ec, the initial electric field inside microchannel ("initial" here means before induced electric field formation by OH- and H3O+ ions in water );

U=2Eb + Ec (Eq.1)
Eb/Ec=εc/εb (Eq.2) (electric field is inversely proportional to dielectric constant)

The results of Eb and Ec are 1630 kv/m, 54 kv/m respectively, in the case of U = 1000 V.

Is this solving method correst?

 Quote by mfb I think you need numbers for the mobility of OH- and H3O+ ions in water, driven by an electric field. If the time-scale there is longer than the time-scale for diffusion*, you are probably fine, otherwise the electric field strength will decrease over time and you need AC which is quicker than that. Keep in mind that the main voltage drop will occur in the dielectric plates anyway. *I played a bit with equations, and think that the equilibrium is an exponential function for the concentration, with a typical length given by the ratio of "influence of the electric field" to diffusion. If this typical length is >>400µm, DC is fine. If it is <<400µm, DC would give nearly no field inside, apart from the boundary regions, between those regions it is something in between.
Regarding diffusion issue, I think your aforementioned "typical length" might be "debye length", which is the distance over which significant charge separation can occur.

I look for the relative info online (wiki and online course note), generally the debye length is in nanometer scale, which is << 400μm.

Could I come to a conclusion that in my experimental setup DC can not give any electric field inside H2O-filled microchannel?
 Voltage is field strength times distance. Thats why the unit for E is V/m. V/m * m = V So U = 2*Eb*hb + Ec*hc The water is conductive. In a static setup - i.e. when no current is flowing and all charges are static - the inside of a conductor is always free of any electric field. When you turn on a DC voltage source you get a current flowing for a tiny fraction of a second and then it stops and the field inside the water has disappeared. How quickly the field in the water disappeares depends again on the capacity and the resistance. The capacity is Cb/2, where Cb is the capacity of a dielectric plate You effectively have 2 capacitors and a conductor in series. With that you can calculate the time constant. http://en.wikipedia.org/wiki/RC_time_constant τ = R*C The time constant is the time in seconds that it takes to fill a capacitor to 63% Diffusion plays no role here except maybe if your dielectric plates are not perfectly watertight and allow ions to pass through. To get a constant field inside the water with DC you need to let a constant current flow through the water. That means the water needs to have direct contact with the metal. The more pure the water the higher it's resistance and the less current you need.

Mentor
 Quote by DrZoidberg Diffusion plays no role here except maybe if your dielectric plates are not perfectly watertight and allow ions to pass through.
Diffusion leads to a non-zero Debye length, and this can lead to electric fields even in the static case. The concentration gradient of the ions leads to an effective force against the electric field.

@0lolol0: Hmm, could be. Pure water has a low concentration of OH- and H3O+, which should give some larger value for the Debye length. But probably not 400µm.
So the main question now is the time-scale of this.

 Quote by mfb Hmm, could be. Pure water has a low concentration of OH- and H3O+, which should give some larger value for the Debye length. But probably not 400µm. So the main question now is the time-scale of this.
Thanks, mfb.

I found a detailed value of Debye length of pure deionized distilled water, it wrote as 1 μm.
http://www.uic.edu/classes/phys/phys450/MARKO/N005.html

I think it would be fine compared with 400 μm distance, right?

Regarding timescale of screening, is there any other way to solve that, besides DrZoidberg's methods based on time constant?

Thanks again.

 Quote by DrZoidberg How quickly the field in the water disappeares depends again on the capacity and the resistance. The capacity is Cb/2, where Cb is the capacity of a dielectric plate You effectively have 2 capacitors and a conductor in series. With that you can calculate the time constant. To get a constant field inside the water with DC you need to let a constant current flow through the water. That means the water needs to have direct contact with the metal. The more pure the water the higher it's resistance and the less current you need.