Specifically, frequencies 5.75GHz and 2.4-2.485GHz.

I want to place a hermetically sealed circuit at the bottom of a gold fish bowl, and want to know if the water will significantly attenuate WiFi or Bluetooth.

Gold fish bowl would be 6" to 1'.

So, I want to know rate of signal absorption per distance through intervening medium of water.

closed as off-topic by knzhou, sammy gerbil, Gert, user36790, John Rennie Jul 22 '16 at 7:28

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    I'm voting to close this question as off-topic because of insufficient research effort. – sammy gerbil Jul 22 '16 at 0:15
  • The only way that we can find absorption rates at these frequencies is by researching. Aren't you able to do your own research? – sammy gerbil Jul 22 '16 at 0:28
up vote 4 down vote accepted

The electromagnetic field in a medium gets attenuated exponentially. $$ \mathbf E = \mathbf E_0e^{-x/\delta} $$

Where $x$ is the distance the signal has traveled. Since the power of the signal is proportional to the square of the field, then the power will be attenuated by $P = P_0 e^{-2x/\delta}$. The quantity $\delta$ is called skin depth. Its the distance that the signal will attenuate its field by $1/e$. So, your question is: what is the skin depth of water at microwave frequency range?

That depends... Depends on the conductivity and dielectric constant of the water. The more salty the water is, the greater the conductivity, and thus, the greater the attenuation.

Let $\epsilon_0\approx 8.812\cdot 10^{-12} F/m$ be the electric permittivity of vacuum, $\mu_0 = 4\pi\cdot 10^{-7} H/m$ the magnetic permeability of vacuum, $\epsilon_r$ the dielectric constant of water (which is approximately $80$), and $g$ the conductivity of water. The skin depth can be calculated with such parameters:

$$ \frac{1}{\delta} = 2\pi f\sqrt{\frac{1}{2}\epsilon_r\epsilon_0\mu_0 \left(-1 + \sqrt{1 + \left(\frac{g}{2\pi f\epsilon_0\epsilon_r}\right)^2}\right)} $$

Notice that, if $g$ is much greater than $2\pi f\epsilon_0\epsilon_r$ in the minimum of 2 orders of magnitude, ie $g\gg 2\pi f\epsilon_0\epsilon_r$ then the skin depth can be approximated: $$ \delta = \sqrt{\frac{1}{\pi fg\mu_0}} $$

If $g\ll 2\pi f\epsilon_0\epsilon_r$ it can be easily seen this approximation: $$ \delta = \frac{2}{g}\sqrt{\frac{\epsilon_0\epsilon_r}{\mu_0}} $$

In the first approximation (good conductor approximation), notice that, as conductivity increases, the skin depth decreases, then attenuation increases. And in the second (bad conductor approximation), as conductivity increases the skin depth decreases then the attenuation increases. Thus, in all cases, the greater the conductivity, the greater the attenuation. According to this data then, we have for water, at a signal with $f = 2.485GHz$:

  • Pure water. $g\approx 5.5\cdot 10^{-6} S/m$. Then, $g\ll 2\pi f\epsilon_0\epsilon_r$. Then $\delta\approx 8.01Km$.

  • Drinkable Water. $g\approx 0.001S/m$. Then, $g\ll 2\pi f\epsilon_0\epsilon_r$. Then $\delta\approx 44.54m$.

  • Sea Water. $g\approx 5S/m$. Then, $g\gg 2\pi f\epsilon_0\epsilon_r$. Then $\delta\approx 8.91mm$.

As you can see, salty water is evil. After signal has penetrated $8.91mm$ inside sea water, its power decreases by factor of $1/e^2\approx 0.13534$ ie $86.466\%$ of its power was lost. In WW2 this was a problem to overcome in submarine communication. =). In pure water, we show the wave needs to travel $8.01Km$ to lose its power by $e^{-2}$ factor. A lot to travel. Thus, won't be too much attenuated. In your case however, I guess its drinkable water. However, assuming the $0.05S/m$ for drinkable water, we get skin depth of $80cm$. So, the skin depth varies too much for little variation in conductance. Thus, you will need to measure the conductivity of your water.


There is also the problem of reflectance. A signal might be reflected back by water. The skin depth relates to water imaginary optical constant $k$ by: $$ \delta = \frac{c}{2\pi fk} $$

The water reflectance, that is, the ratio of the power that is reflected, at a normal incidence angle, can be calculated: $$ R = \frac{(n-1)^2 + k^2}{(n+1)^2 + k^2} $$

Where, $n$ is the refractive index of water, that is, $n\approx 8.5$ for this frequency. The relation $n = \sqrt{\epsilon_r}$ might help. Since $k$ is calculable and related to the $\delta$, one can show that, the greater the attenuation (low skin depth), the greater the reflectance. Thus, you have two problems to overcome: The reflectance of the interface air-water, and about the non-reflected waves, ie, the ones inside the water, there is the skin depth (attenuation).

Be aware that this reflectance is only for normal incidence. For incidences with a generic angle $\theta$ (ray cast), the reflectance will change (often to increase its value, ie, reflect more).


For drinkable water (probably your case (there is a fish involved...)), then we have at worse scenario, skin depth of $\delta\approx 80cm$ and reflectance of $R = 0.61940$, meaning roughly $62\%$ of incoming radiant power at normal incidence will be reflected back, and won't penetrate the water. Only roughly $38\%$ will penetrate water, and encounter exponential attenuation of $e^{-2x/0.8} = e^{-x/0.4}$.

Now.. your final formula! Assuming signal leaves antenna with power $P_0$, the amount of power $P$ that the receiver will receive is: $$ P = \frac{P_0}{4\pi r^2}(1-R)e^{-x/\delta},\quad\quad P = \frac{P_0}{4\pi r^2}0.38060 e^{-x/0.4} $$

Where $r$ is the distance between the antenna and the receiver (assuming spherical propagation), and $x$ is the distance that the signal will have to travel thru water.


Out of note, there is no such thing as "rate of signal absorption per distance through intervening medium of water" as you suggested, because the attenuation is exponential, as opposed to linear.

Well, hopefully, besides answering your question, I hope I helped you learn something new of physics. =).

  • Nice answer. +10 from me. :-) – Gert Jul 22 '16 at 0:19
  • This is why I still love physics at age 59. It is so understandable! And, how cool that it lets you predict and manipulate the future?! – Doug Null Jul 22 '16 at 20:25
  • Also, thanks for comparing pure vs. drinking water. I'll use pure! (and a fake fish) – Doug Null Jul 22 '16 at 20:29

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