# How dangerous is a parallel plate capacitor which is used in physics demo experiments?

How dangerous may it be when experimenting with a parallel plate capacitor typically used in physics demonstration experiments? Take for example this one (that's what I use): http://www.ld-didactic.de/ga/5/544/54422/54422de.pdf

Depending on the plate distance the capacity varies. I tried to estimate if this could be dangerous but I am not sure about it. Here is what I did:

From the manual, the capacity $C$ for a plate distance $1\,\mathrm{mm}$ is about $C = 550\,\mathrm{pF}$. Suppose you charge it using $U_0 = 25\,\mathrm{kV}$. And suppose your fingers are a bit sweaty, thus I assume a body resistance of $R = 500\,\mathrm{\Omega}$.

The discharging equation is:

$$I(t) = \frac{U_0}{R} \mathrm{e}^{-\frac{t}{RC}}$$

This gives me an initial current $I_0 = I(0\,\mathrm{s}) = \frac{U_0}{R} = 50\,\mathrm{A}$ which would be extremely dangerous if applied over a certain time. However in a very short time, in $10^{-6}\,\mathrm{s}$ the current drops below $0,02\,\mathrm{A}$, which should be save.

I am not quite sure if my considerations are correct and appropriate and I am not sure which conclusion I should draw, how dangerous it is (or not).

I guess that it doesn't make much difference in this case if $2\,\mathrm{kV}$ or $25\,\mathrm{kV}$ were used. The key factor seems to be the time in this case (because in each case the initial currents are extremely dangerous if applied long enough, so the question seems to be: is the time short enough to make no harm?)

Edit

If you charge it say with $25\,\mathrm{kV}$ at $1\,\mathrm{mm}$ and then enlarge the space to the maximum i.e. to $70\,\mathrm{mm}$, the voltage would potentially raise to $1750\,\mathrm{kV}$, however the capacitance would drop to $7,8\,\mathrm{pF}$. Would doing this operation change anything significantly concerning the safety of the experiment?

Would be great to get a detailed and comprehensive analysis of the safety of such a kind of capacitor.

In your case, $$E = \frac{1}{2}C V^2 = \frac{1}{2}\cdot 550\cdot 10^{-12}\cdot (12\cdot 10^3)^2\,\mathrm{J} = 0.04\,\mathrm{J}$$ a mild static shock.