# How can we create static electricity with high voltage DC without a Van der Graaf generator?

I saw some videos on Youtube like this one from "Join the technicians" and this one from "Electroboom". The videos show an application of static electricity. In school, we were taught that we can generate static electricity with a triboelectric effect with a Van der Graaf generator. However, in those videos, they all use voltage multipliers to bump up the DC voltage super high, without any moving parts. Can anyone explain how these things work? I understand how a Van der Graaf generator works. Those circuits seem nice, I mean you use a lot of parts from a Van de Graaf generator to create static electricity, some wear out fast like the motor or the belt. If high voltage DC just like that can be used, why don't we just use circuits like those?

Thank you in advance!

• Cockcroft–Walton generator and the modern equivalents. – Farcher Aug 10 '20 at 10:09
• As someone who has routinely used both VdG and C-W ion accelerators over the years, your intuition of their reliability is incorrect. Sure, anything below 300kV and a C-W is great. At several MV, not so much. My 6MV on terminal VdG has gone >5 years between maintenance in the tank. – Jon Custer Aug 10 '20 at 12:26

## 1 Answer

There is a fairly simple way of generating a large static charge from a high voltage. You need two brass (or other metal) plates, and a very thin insulating plate with fairly high permeability, 0.5mm teflon is good (but 0.5mm lexan is easier to find), and thick enough to withstand the DC high voltage; some jumper leads; a long insulating handle attached to one of the plates (length depends on how much voltage you get); and a brass ball (or whatever you want to store your charge onto) on an insulated pole (or whatever).

1. Simply take one plate (without the handle) , place on a table or similar. attach it to ground (i.e. ground pin on a power point or a piece of metal piping) and one of the power supply leads (assuming a floating supply),
2. Place the insulator sheet over the plate,
3. Place the other plate (with handle) on top of this,
4. Briefly touch a jumper lead from the HV supply to the top plate to charge it,
5. Lift off the plate and touch it to the brass ball to transfer some charge,
6. Repeat about 4 times (the ball gets 1/2, then 3/4, then 7/8, then 15/16 of maximum charge at each step).

Apparatus diagram: What happens is you are making a capacitor C, ( $$\epsilon_0 = 8.85pF/m$$, $$A$$ = plate area in $$m^2$$, $$d$$ = plate seperation in m ) $$C = \epsilon_r . \epsilon_0 . \frac {A}{d}$$ With plates that are 150mm diameter (0.018sqm) , and using a 0.5mm lexan insulator $$(\epsilon _r = 3)$$ then $$C = 3 * 8.85 * \frac {0.018} {0.0005} = 938pF$$

You then charge it up with a charge $$Q = C . V = \epsilon _r .\epsilon _0. \frac {A}{d} * V_{HV}$$ When you pull the plate away , the charge stays the same, but the capacitance drops significantly; in order to conserve charge the voltage must increase proportionally $$C_1*V_1=Q=C_2*V_2$$; simply put: reducing capacitance to one tenth increases the voltage tenfold .

So separating the plates a distance of 150mm ($$\epsilon_r = 1$$ for air) lowers the capacitance to $$C= 1 * 8.85 * \frac {0.018} {0.15} = 1.04pF$$ Note that the capacitance has apparently dropped by a factor of 900, However we need to consider the disc has self capacitance as well , $$C_s= 8 \epsilon_0 * r = 8 * 8.85 * 0.075 = 5.31pF$$, so we need to use this value once the plate is "sufficiently far away", this yields a capacitance ratio of $$\frac{C_{near}}{C_{far}}= \frac{938}{5.31}=176$$ so if your charging supply is 10 x 9volt batteries = 90v , then the voltage on the plate (when pulled away from the charging station) will be 176 x 90v = 15,900v. You will get proportionally higher voltages by making your dielectric thinner, e.g. a 50um varnish coat will see a tenfold increase.

The resulting voltage of your storage device , when touched with the charged plate, depends on the self capacitance of the storage device, respectively this is Disc: $$C_d = 8 * \epsilon_0 * r$$ , sphere $$C_d = 4 \pi * \epsilon_0 * r$$ , where r is in metres, so a 100mm diam ball is 0.075m radius, hence $$C_{ball} =5.56pF$$ As the two capacitances are similar, if your disc is first charged to 16kV , then touches a ball with no charge then the full charge is split over the two, and the voltage will average out to 8kV on each, the next cycle will have 12kV , then 14kV, then 15kV.

You could use a human volunteer as the charge storage device, with a capacitance of ~40pF, it would take many cycles to get a significant charge on to them, they should be holding a brass rod (to touch the charged disc to) to minimise the yelp factor.

You could automate the whole process by gluing the "movable" metal sheets on a perspex wheel and spinning it around, but if go to that much trouble you might as well build a Wimshurst machine, as it doesn't need the DC supply! https://en.wikipedia.org/wiki/Wimshurst_machine

see also:

• Thank you for the comprehensive answer! I'll look at this, although I actually want to build a permanent, automated one. I'll look at Wimshurst. – gregory112 Aug 12 '20 at 5:57
• You can also use Voltage multipliers (Like cockroft walton style with an AC input) or Marx Banks. The problem with these is they can produce a lethal amount of energy. Normally for "static electricity" you would want something that just stored enough charge to make a nice arc, but only enough charge to produce a jolt if accidentally discharged through a human. – BobT Aug 14 '20 at 13:41