Static electricity: High voltage but low energy

Question

I watched a video of a serie called "The mechanical universe" recorded at California institute of technology in which was said that

"A Van der Graaff generator with a voltage of near 100.000 volts only stores an amount of energy of 2.000 joules while a common battery of 9 volts stores 20.000 joules".

I think of potential as an indicator of energy since it represents the potential energy of the charge unit at a point. (Ep=V*q). If V increases, Ep increases. ¿What is wrong here? In addition, XVIII century high voltages machines are used to teach electricity with experiments in which the current bristels hair of people.. Why so high voltage does not produce a danger current on human body?. I think that from a given value of voltage, current depends only on resistence. Why these machines produce always a low current. ¿What is the mistery of static electricity?

Answer 1

Why so high voltage does not produce a danger current on human body?. I think that from a given value of voltage, current depends only on resistance. Why these machines produce always a low current. What is the mystery of static electricity?

First of all, it is not voltage that causes dangerous electric shock (e.g., ventricular fibrillation). It is a combination of the magnitude and duration of the current. And while the voltage delivered by the Van de Graaff generator you describe may be very high, the current it delivers to the body lasts only an instant. (Very high energy generators, on the other hand, may be dangerous).

The Van de Graaff generator is basically a charged capacitor. The energy stored in the electric field of a capacitor is given by

$$E=\frac{1}{2}CV^2$$

So the capacitance required to store 2 J at 100,000 volts is 4 pF, or 4 x 10$^{-12}$F. Now consider the current that the generator can deliver to the body. According to IEC 60479-1, the internal body impedance is on the order of 500 $\Omega$ hand to hand or hand to foot. The discharge current delivered to a resistor by a capacitor charged to voltage $V$ as a function of time is

$$i(t)=\frac{V}{R}e^{\frac{t}{RC}}$$

For 100,000 Volts and body resistance of 500 $\Omega$ we have.

$$i(t)=200 e^{\frac{t}{RC}}$$

In the equation, $RC$ is the time constant, or the time it takes for the current to be roughly 37 % of its initial value. In this case the time constant is 2 nanoseconds, or 2 x 10$^{-9}$s

So in 2 nanoseconds, the current falls to 74 amps. In 50 microseconds it falls to 4 x 10$^{-20}$A. In 1 microsecond its so small I get an error on my calculator.

The point is, these combinations of current and time, while may be sensed, are far below the thresholds of dangerous electric shock, per IEC 60479-1.

Hope this helps.

Answer 2

Ep=V*q

V is huge but q is limited to what was stored. In a battery, carriers are produced along its life as consequence of the internal chemical reaction.

But anyway, if static electricity is safe or not, it depends on the amount of charge present. As an extreme example, a lightning results from discharge between clouds, or between clouds and earth, and can be really dangerous.