What's the electric field intensity inside parallel-plate capacitor filled with water?

Question

                                   _______________________
                                  |                       |
           ---------------------------                    |
            copper plate                                  |
           ---------------------------                    |
            PDMS plate                                    |
           ---------------------------                 ======= 
            Microchannel filled                          === 500 DC V
            with H2O                E?                    |
           ---------------------------                    |
            PDMS plate                                    | 
           ---------------------------                    |
            copper plate                                  |
           ---------------------------                    |
                                   |______________________|
            Cross-section view

I'm sorry for your inconvenience to see the above diagram, since my reputation is too low to insert image file. >_<

Thinkness: copper=300um; PDMS(polymer)=300um; Microchannel depth=400um;

Dielectric constants of PDMS and H2O are 2.65 and 80, respectively.

My experimental purpose is to generate uniform electric field inside of water-filled microchannel by appling electric field externally. (“externally” here emphasizes no direct contact between metal electrodes and water; here PDMS, a dielectric polymer, serves as a barrier between them)

Question 1: If DC V = 500 V; then what is electric filed intensity inside of water-filled microchannel? 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, the final electric field intensity inside water-filled microchannel would go to zero after instantaneous screening process?

Question 2: If alternating current voltage, like sinusoidal AC (100 kHz, peak voltages from 0 and 500 V, the bottom copper plate is grounded), is applied instead of DC, what would be electric field intensity inside microchannel? Taken electrostatic screening issue into account, what the minimal frequency of ac should I set?

Many thanks in advance for your attention to this matter.

Answer 1

About the autoionization of water ...

Wikipedia (http://en.wikipedia.org/wiki/Debye_length) gives a formula for water

$$\text{debye length in nm} = \frac{0.304}{\sqrt{I\text{ in molar}}}$$ where $I$ is ionic strength, which is 1E-7 for pure, pH-neutral water. That gives a screening length of 1$\mu$m.

So at DC, there will be an electric field in the bottom 1 micron and top 1 micron of the water, with no electric field in the central 99% of the water. The field will be canceled by the OH- ions spread near the top electrode and the H+ ions spread near the bottom electrode.

Despite what you say, this screening process is NOT instantaneous. The H+'s and OH-'s have to travel to the appropriate side. Therefore at AC the screening may be much less or completely negligible. To calculate the frequency required, you would do a calculation involving the electrical resistance of pure, pH-neutral water, the initial voltage pulling the charges, the quantity of charge that has to move from one side to the other, and the distance that it has to move. I don't have time for this part and don't have any intuition for what order of magnitude of frequency would be the cutoff between screening and non-screening behavior.

Answer 2

1) First of all, screening is a response to an external field, and as such it can never fully counteract the effect, except in the case of a perfect conductor. As $\varepsilon_\text{water}$ is finite, we know we are not in the perfect conductor limit. If an external field is applied, charges will rearrange (via bulk motion of ions, re-orientation of polar molecules, and/or induced polarity in neutral atoms/molecules) to partially cancel the field. Gauss's Law, together with appropriate geometry considerations, tells us the electric field for fixed surface charge density $\sigma$ is $E = \sigma/\varepsilon$.

Of course, you aren't fixing charge density, but rather voltage. To find the charge density from the voltage, you need the capacitance of the system: $Q = CV$. This is a series of three capacitors, and so the capacitance of the system is the reciprocal of the sum of the reciprocals of the individual capacitances, which are $C_i = \varepsilon_i A/d_i$ for ideal parallel-plate capactors.

If there were only one material, you'd find the effect of $\varepsilon$ on capacitance is exactly balanced by its reduction of the electric field for a given charge density, and so $E = V/d$ no matter the material. Your problem is slightly more complicated, but in the end you should find a nonzero field strength in the water.

(Note: Everything here is in the limit of instantaneous transitions. Not being an experimentalist, I don't know off hand the length scale over which screening takes place. The problem is somewhat more complicated if that length is comparable to the thicknesses in your setup, since then step functions are replaced by exponential attenuation.)

2) You can use AC. Then you must take into account the frequency-dependence of $\varepsilon$ for each material. A very rough rule of thumb in most typical situations is that $\varepsilon$ is reduced at higher frequencies, since the charges have less time to rearrange by whatever mechanism. However, there are resonances that need to be taken into account.

Answer 3

If I remember correctly from Electromagnetism, if you wait for the circuit to reach equilibrium with the capacitor (i.e. current = 0) doesn't the total voltage across the entire circuit have to be zero? So the if $\Delta V_{capacitor} \approx 500V$ and since in a parallel plate capacitor the electric field is relatively constant. So you have: $$\Delta V = EL = 500,$$ $$E = 500/L$$ where $E$ is the electric field and $L$ is the distance from one plate to the other?

If your hypothesis in Question 1 was correct and the electric field was zero, then there would never be any voltage going against the battery. Constant current = constant build-up of charge on the plates, I don't think that's likely.