Suppose you have a battery, and using some wire you connect the positive terminal of the battery to some conductive piece of metal (i.e. an electrode) and similarly you connect the negative terminal of the battery to another electrode. Now place both of these electrodes into water. Using a voltmeter, if I then probe the voltage $\phi$ at various points within the water (relative to the negative terminal of the battery), I actually get a non-zero reading: enter image description here

This suggests to me that an electric field $\textbf E$ exists within the water, since: $$\textbf E=-\vec{\nabla}\phi$$ However, I'm struggling to understand qualitatively where this electric field is coming from. I know from Maxwell's equations that electric fields can only arise from distributions of electric charge or from time-varying magnetic fields. However, the fact that the voltmeter produces a well-defined voltage reading $\phi$ suggests that: $$\vec{\nabla}\times\textbf E=-\vec{\nabla}\times\vec{\nabla}\phi=\textbf 0=-\frac{\partial\textbf B}{\partial t}$$ So there shouldn't be any time-varying magnetic fields. This means that $\textbf E$ is arising from some sort of electric charge distribution. At first glance, this seems quite reasonable since the experimental electric field that I obtain using the voltmeter looks quite similar to the electric field of an electric dipole: enter image description here

However, as soon as I switch the liquid from water (which is polar) to oil (which is non-polar), keeping everything else the same, the electric field suddenly vanishes $\textbf E=\textbf 0$. Similar experiments with other liquids like milk and bleach revealed to me that the liquid's polarity seems to play a decisive role in determining whether or not an electric field $\textbf E$ will appear. So I have a couple of questions:

$1.$ If the electric field $\textbf E$ is supposedly arising from an electric charge distribution, where is this electric charge distribution even coming from? From electrons coming from the negative terminal of the battery? From an electrolytic process within the water?

$2.$ Why does this have anything to do with the polarity of the liquid?

$3.$ Does some kind of electric current get formed in the water? What is the nature of this electric current? What role would such an electric current play in generating an electric field $\textbf E$?

  • $\begingroup$ Is it true that “as soon as I switch the liquid from water to oil, the electric field suddenly vanishes”? I think that is incorrect. Do you have some experiment that confirms that? $\endgroup$
    – Dale
    Commented Apr 30, 2022 at 12:53
  • $\begingroup$ Do you understand why there is a potential difference between different points along a solid conductor with DC current? There is no variable magnetic field either. Do you think the liquid conductor should behave differently in this respect? $\endgroup$
    – nasu
    Commented Apr 30, 2022 at 13:56
  • $\begingroup$ Is your water tap water, or has it been purified via deionization or distillation? It might be worth trying your experiment with purer water to see if the effect persists. $\endgroup$ Commented Apr 30, 2022 at 14:14
  • $\begingroup$ @Dale Yes it is true, I did experiments and that is simply what I observed. $\endgroup$ Commented Apr 30, 2022 at 22:13
  • $\begingroup$ @MichaelSeifert The experiment works equally well with tap water as with distilled water (in the sense that in both cases, the voltmeter gives non-zero readings all over the liquid, so an electric field $\textbf E$ exists in both cases). $\endgroup$ Commented Apr 30, 2022 at 22:17

3 Answers 3


The problem lies in the measurement method, not in the medium itself. As soon as you connect the battery, you will always have an electric field (actually also if you do not connect the battery), no matter what the medium is (oil/water/air), because the battery has a voltage, which inevitably extends into the rest of space according to Maxwell's equations.

The point is that it is often not as easy to measure this electric field, because the measurement device disturbs the original field. Suppose you are using an ordinary digital voltmeter to measure the potential distribution, and you are measuring in oil or in air or any other electric insulator. Then this voltmeter has a typical inner resistance in the 10's of Megaohms. The two terminals of the voltmeter act as a capacitor which is charged up against the voltage to measure, and the corresponding current flows over the inner resistance of the voltmeter. You are only able to measure the voltage accurately as long as that virtual capacitor is not charged up to the maximum, because as soon as it is charged up, current stops flowing and the voltmeter will measure ~0 Volts. You might think that a resistance of 10 MOhm is huge, but the capacitance of the terminals is even way smaller (I guess in the picofarads), so that it takes almost no time to charge the capacitor. You would probably be better of using an oscilloscope because then you could trace the charging by the time evolution of voltage.

Now, if you are using water to measure the potential distribution, you have to know that water is not only polar (which increases the capacitance of the virtual capacitor a little bit), but particularly tap water also contains tiny amounts of salts which make the water conductive. So, the conductivity of the water ensures a steady supply of electric current from the battery, some of which flows through the voltmeter and allows the continued measurement of the potential. As long as the interior resistance of the voltmeter is much higher than that of the water, the measurement will work. You may ask, what happened to the virtual capacitor. Well it is still there, but it is in parallel to the resistance of the water, so it is unable to stop the flow of current.

I would speculate that you would obtain almost the same result as for oil/air if you use demineralized water instead of tap water, because then water does also behave like an insulator. The nominal increase in capacitance due to water polarizability will not be noticeable with a digital voltmeter. You will again just measure zero. Again, by using a good oscilloscope you might be able to detect the change in capacitance between air/oil/(demineralized) water.

One interesting aside: high voltages can be measured by an electroscope. It works by exploiting the inevitable capacitance of the measurement arrangement to our advantage. The capacitor consists of a needle against the housing of the device. If the capacitor is charged, the needle moves due to electrostatic forces. Therefore, the electroscope does not need a steady flow of current for being able to measure. Hence, also the interior resistance is almost infinite. But, it only works for high voltages because of the relatively high forces needed for the needle to move. I guess, making the arrangement smaller also wouldn't help for lower voltages, because that also decreases the capacitance.

  • $\begingroup$ Unfortunately your speculation is incorrect, as I do get good readings on the voltmeter for both tap and distilled water, but not for oil. This is what leads me to suspect that the liquid's polarity is the key determinant of whether or not there will be an electric field $\textbf E$ in the liquid. $\endgroup$ Commented Apr 30, 2022 at 23:29
  • $\begingroup$ @SurfaceIntegral Even perfectly pure water is only 18 MOhm cm. Commercial distilled water is more like 0.5 MOhm cm. Oliver is correct in principle regarding the experimental issue, but any water is simply conductive for this purpose $\endgroup$
    – Dale
    Commented May 1, 2022 at 14:37

Water conducts electricity, so the battery drives a steady current through the water. There is therefore a current desity $$ {\bf j}= \sigma {\bf E}= - \sigma \nabla \phi$ $$ where $\sigma$ is the conductivity. In a steady state we have $\nabla \cdot {\bf j}=0$ so $\nabla^2 \phi=0$. The voltage distribution in the water is therefore the same as it would be if there were charges at the electrodes rather than a current source.

I am surprised that your setup works so well, as there are often all sorts of confounding chemical effects from the surfaces of the electrodes, and also you need a high-restance voltmeter.

  • 1
    $\begingroup$ The OP notes that if they switch from water to oil the effect vanishes, so perhaps their setup is not as robust as all that. It might be worth editing your answer to discuss why the voltmeter would read zero if the resistivity of the liquid was too high. $\endgroup$ Commented Apr 30, 2022 at 14:11

You have omitted the effects of electrochemistry, by which the application of a voltage difference between two pieces of metal facing one another in water will cause the formation of ions in the water which let it conduct electricity.

And if there are pre-existing ions already in the water (like dissolved salt) then this process can go forward even if the metal electrodes to not create metal ions themselves. Even very small amounts of (for example) salt in the water will have a very significant effect, causing the water to become conductive.

  • $\begingroup$ However, the experiment worked even when I used distilled water (in the sense that I still observed a non-vanishing electric field $\textbf E$). Therefore, I suspect that it is the polarity of the liquid that matters most, rather than the liquid's conductivity. $\endgroup$ Commented Apr 30, 2022 at 22:43
  • $\begingroup$ distilled water will read 11 megohms simply because of the natural dissociation of the water molecules and has nothing to do with the polarity of the molecules. $\endgroup$ Commented May 1, 2022 at 1:00

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