This question already has an answer here:

If a person has plastic soles, and are playing on Artificial grass and then the person touches metal goal post, the person will get a static shock.. Why?


marked as duplicate by John Rennie, Daniel Griscom, user36790, Gert, Ali Jan 12 '16 at 6:55

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Hello, and welcome to Physics SE. Look around and take the tour. As your question stands, it does not show much effort on your part. A search on this site, or on Google, would lead you to useful information. $\endgroup$ – Jon Custer Jan 11 '16 at 16:56
  • 4
    $\begingroup$ Possible duplicate of Help me understand static electricity $\endgroup$ – Jon Custer Jan 11 '16 at 16:57
  • $\begingroup$ @JonCuster the duplicate doesn't really address how you can discharge to a metal object that is not (DC, low impedance) "grounded". Which is an important aspect of the question. I believe there are other duplicates on the site though... I may have time to take a look later. $\endgroup$ – Floris Jan 11 '16 at 17:56

There are two aspects to your question. One is: "when I run across an artificial turf field with plastic soles, why do I accumulate static charge?" and the second is "When I am charged, why do I experience a shock when I touch a metal goal?".

Taking each in turn:

There is a phenomenon called triboelectricity - the generating of a potential difference when two dissimilar materials rub together. In essence, during the rubbing electrons are briefly "knocked loose"; for dissimilar materials, they have different affinity for the different surfaces, and therefore there will end up being a net charge difference between the surfaces after rubbing (more details in the link provided). Now this charge will dissipate into the rest of your body - after all, your body is a pretty good conductor, and the charge that's bunched up on the sole of your foot would like to be distributed over a larger area (since like charges repel). As the charge dissipates, this creates an opportunity for more rubbing to lead to further charge transfer, until equilibrium is reached. In the equilibrium situation, the potential on your body is such that the rate at which you lose charge to the air around you (a function of humidity, mostly) is equal to the rate at which you charge up (a function of the potential you already have, and the amount of shuffling you do). Incidentally this is the reason that you get charged to a higher voltage on cold days - relative humidity is lower when it's cold outside. More details in this earlier answer

Now the second part - why do you get a shock when you touch the goal? After all, there may not be a closed circuit - you are not standing on a metal surface connected to the metal goal. To understand this we need to invoke capacitance. This is simplest to explain if we assume humans are spheres - the old "spherical cow" assumption of physicists... The capacitance of a conducting sphere is given by $$C = 4\pi\epsilon_0 R$$

Further, since $Q=CV$, once we know the voltage and the capacitance, we know the charge. Assuming $R=0.4~\rm{m}$ and $V=20~\rm{kV}$, we find the charge is approximately 1 µC. Now if the goal is initially uncharged, and not connected to ground, it still has a capacitance. If we assume that the effective capacitance of the goal is half that of the person (there is more length, but the radius is less - I just want to demonstrate the principle here) then when you touch the charge will redistribute until the voltage is the same on both you and the goal. This means that the final charge distribution will be in proportion to the capacitance, so you end up with $\frac23$ of the charge on the person, and $\frac13$ on the goal (when the ratio of capacitances is 2:1, the charge must be in the same proportion). This means that about 300 nC of charge flows. If the charge flows in 1 µs, the mean current during the spark would be 0.3 A - but the total energy is a very small fraction of a Joule. So you feel a shock, but it doesn't harm you.

See also this answer (I used a slightly different value of $R$ there... the principles are the same).


Not the answer you're looking for? Browse other questions tagged or ask your own question.