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(details as per link below)

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

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- Thread starter Alex Schaller
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In summary, if you have a conducting sphere with a charge, the electric potential outside of its surface is V=-∫E→⋅d→r. The breakdown field value iEmax value will give a relationship between ##r_{min}## and V (or Q if you wish).f

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(details as per link below)

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

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What do you think? And why?

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

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

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Do you agree, so far?

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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

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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?

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

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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##?

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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).

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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 iEProvided 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?

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