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Water is a polar molecule, which means it has an uneven charge distribution. How much of an electric field strength would it take to align all, or most, of the molecules at room temperature?

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  • $\begingroup$ Pure water, aye? $\endgroup$ – BMS Jan 20 '15 at 21:45
  • $\begingroup$ Microwave ovens work because water molecules (and other polar molecules) try to align with the rapidly oscillating microwave field...so I guess it doesn't take that much field strength. $\endgroup$ – Nathan Reed Jan 20 '15 at 21:48
  • $\begingroup$ Or maybe not much alignment, but lots of it done fast $\endgroup$ – user56903 Jan 20 '15 at 22:05
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You get a rough estimate by comparing the energy lost by the dipole as it aligns with the field $\sim d E$ with the thermal energy $\sim k_B T$. Only when the former is much larger than the latter will the alignment be stable against thermal fluctuations. Taking this website's value for the dipole moment of water $$ d \approx 1.85 D \approx 1.85 \times 3.34\times 10^{-30} C m, $$ and $T = 300 K$, I get that $$E \sim \frac{k_B T}{d} = 6.7 \times 10^{8} Vm^{-1}.$$ I would count that as a pretty strong field.

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    $\begingroup$ Not something easily done with a DC circuit, but quite do-able within a laser pulse $\endgroup$ – user56903 Jan 20 '15 at 22:18
  • $\begingroup$ @DirkBruere True, but I am not convinced that a laser pulse will produce any meaningful alignment, because the field only points in a well-defined direction for less than an optical period, which is on the order of $10^{-14}$s. $\endgroup$ – Mark Mitchison Jan 22 '15 at 3:58

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