It is often instructive to see a practical application of theoretical principles, so here are some examples:
One limit on the maximum voltage that a Van deGraaff generator can reach is when the electric field strength at the surface of the sphere exceeds the breakdown for the insulating gas it is immersed in. The section on Van deGraaff generators describes other limitations such as arcing down the belt. For small demonstration type generators operating in room air, however, the E field limit is usually applicable.
Field at surface of a sphere = Voltage / Radius of Sphere
So, we use the handy approximation of 30 kV/cm for the breakdown strength of air and rearrange the equation a bit to get:
MaxVoltage = 30 kV * Radius(cm)
The classic Franklin Lee designed Van deGraaff from Scientific American has an upper electrode about 14 inches in diameter which actually has a minimum radius of curvature of about 7 inches (it is a "squashed" sphere).
MaxVoltage = 30 kV/cm * 17.5 cm = 525 kV
The generator is advertised as reaching 400 kV, which is probably a realistic assessment, given that it isn't perfectly smooth and that there is an insulating column between the sphereoid and the ground plane at the base.
While the upper electrode on that Van deGraaff isn't perfectly spherical, we can probably approximate it as one to calculate the capacitance and stored energy. Note that here, we'll use the real radius of the sphereoid, as opposed to the breakdown voltage calculation, where we used the smallest radius of curvature.
r = 7 inches = .178 meters
Csphere = 4*pi*epsilon*r = 111.2 pF/meter * .178 meters = 19.8 pF
Now, let's calculate the stored energy: Is it a crack or a bang when we discharge it?
Energy = 0.5 * C * V^2
Energy = 0.5 * 19.8E-12 * 4E5^2 = 1.5 Joules
(A handy form of the energy equation is: Energy (Joules)= 1/2 * C(microfarads) * V(kilovolts) ^2. The units nicely cancel.)
Pretty much the first thing people do when they get a demonstration Van deGraaff is to make sparks from the upper electrode. When I did this, I was amazed at the electrostatic force on the grounded electrode I was discharging to. It is one thing to intellectually know about Columb forces, and another to feel them in your arm muscles (or to watch them move the electrodes around). As it happens, calculating the field between two conducting spheres is a non trivial task because the charge isn't evenly distributed. However, let's approximate (since experiment will determine the exact number, we only need to know if it is a tiny force or a huge one).
At 400 kV, the gap between the spheres will be a minimum of 10-15 cm. Let's use 15 cm spacing of 15 cm radius spheres for the calculation. A first order approximation will use Coulomb's law, and treat the sphere as a point charge at the center. First question: How much charge?
q = CV = 19.8E-12 * 400E3 = 7.920E-6 Coulombs
I've approximated C as a sphere in free space, which certainly isn't the case, but will do for an order of magnitude calculation
Now, let's use Coulomb's law
F = 1/(4*pi*epsilon) *q1 * q2 / r^2 = 1/111.2E-12 * 7.92E-6 * 7.92E-6 / (.45^2)
= 2.78 Newtons (approximately)
About a half a pound, which is more than you would expect at first glance.
Copyright 1997, Jim Lux / 14 Sep 1997 / statcalc.htm / Back to HV Home / Back to home page / Mail to Jim