Electric arc length in relation to voltage

[Alex Schaller]

Provided the length of the arc of an electrostatic generator is 7 cm, can we state that its voltage is around 210,000 V?
(details as per link below)

https://photos.app.goo.gl/MKSpviQwPh9eS5jZ7

 

[phinds]

What do you think? And why?

 

[Alex Schaller]

I also know that the sphere's diameter is 30cm. So, its capacitance should be 2.5 pF.

 

[Alex Schaller]

So I think that, in order to hold a voltage of 210,000 V, its original charge should be something around 525 nC.

 

[Alex Schaller]

Do you agree, so far?

 

[tech99]

I don't think the capacitance is anything to do with it. My understanding is that an air gap tends to break down at approx 20 to 30 kV/cm. But it is only approximate, and in his experiments with radio waves, Hertz investigated spark lengths and found that a spark will never occur below about 300 Volts. A spark and an arc are different things, and an arc can be maintained at a low voltage.

 

[Alex Schaller]

The greater the capacitance, the more charge the sphere can hold. The more charge it holds, the greater the electric field it produces. The greater the electric field, the longer the spark. Does this make any sense?

 

[hutchphd]

The greater the capacitance, the more charge the sphere can hold. The more charge it holds, the greater the electric field it produces. The greater the electric field, the longer the spark. Does this make any sense?

You have it wrong I think. It is the maximum electric field strength that initiates the arc, not the total voltage. To maximize the length and strength of the arc one wishes to maximize the charge while minimizing the maximum electric field. You want a big sphere.

 

[Alex Schaller]

A big sphere means it can hold a greater capacitance (and therefore a greater charge, as capacitance is the charge per unit voltage). Could we say this is correct?

 

[Vanadium 50]

No, it is not. Capacitance is charge per unit voltage, not volume. Further a capacitor needs two ends - charge is conserved, so if you put positive charge on one place you need a negative charge somehwere else.

It is true that a 1m sphere has 10x the capacitance of a 10 cm sphere. But it has the same capacitance and 1% of the weight of ten such spheres in parallel. Going big makes this difficult., expensive and hazardous.

The fact you didn't answer the question is concerning. People will conclude your plan is to build a perpetual motion machine or similar. So what is it this gizmo is supposed to do?

 

[Alex Schaller]

It is true that a capacitor needs two ends; a sphere can also be considered a capacitor, where the other end are the walls of the room that encircle the sphere.

Everybody knows that a perpetual motion machine cannot be built (as per the second law of thermodynamics).

 

[Alex Schaller]

I don't think the capacitance is anything to do with it. My understanding is that an air gap tends to break down at approx 20 to 30 kV/cm. But it is only approximate, and in his experiments with radio waves, Hertz investigated spark lengths and found that a spark will never occur below about 300 Volts. A spark and an arc are different things, and an arc can be maintained at a low voltage.

Provided the air gap breaks down at approx 30 kV/cm (which is a magnitude of Electric Field) as you mentioned, couldn't we derive from there the voltage as: $V=-\int \vec E\cdot\vec dr$ substituting $\vec E ~by~ 3000000~ \hat r~ \frac V m$?

 

[Alex Schaller]

If so, would it be proper to replace $\vec dr \$ by $dr . \hat r\$ so as to perform the scalar (dot) product of the integral? (@Vanadium 50 is welcome to answer, although this is more like a math scheme instead of a physics intrigue).

 

[hutchphd]

Provided the air gap breaks down at approx 30 kV/cm (which is a magnitude of Electric Field) as you mentioned, couldn't we derive from there the voltage as: V=−∫E→⋅d→r substituting E→ by 3000000 r^ Vm?

Write down electric potential V everywhere outside the surface of a conducting sphere r with charge Q. This is in every Physics textbook. The radial E field is $$-\partial V / \partial r $$ This breakdown field value iE_max value will give a relationship between $r_{min}$ and V (or Q if you wish).$$E_{max}=\frac V r_{min}$$