Recent content by 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).

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##?

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).

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?

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?

Do you agree, so far?

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

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

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

Indefinite integrals can be regarded as a set (family) of curves, and each of the curves can be obtained by shifting in a parallel the curve, upwards or downwards (along the "Y" axis).

You are welcome

The circuit resembles an EMF connected in parallel with a capacitor, which is then connected also in parallel with a resistor. The voltage on the capacitor is equal to the applied EMF voltage. The capacitor resembles a parallel plate capacitor. Its capacitance (per unit length) is readily...

If the cones are within a conductor, they would contain no (unbalanced) charge; all the charges would try to separate as much as they can (because they are repelling each other) until they reach the surface of the conductor (here they cannot proceed farther as the air surrounding the conductor...

Feynman puts it simple, in his "Lectures on Physics", page 5-5: If the spherical shell is conductive, charge would spread evenly. At any point within the sphere, take for instance "P", the field is due to the contribution of the charges on the sphere. If you think of two opposite cones at "P"...