mutual capacitance?

Oh dear, I thought 1 cm = 1.12 pF? the capacitance of a 1 cm sphere.

George h.

Yup, I got it backwards.

Cheers

Phil Hobbs

Then infinity represents the reference point, at potential zero.

The problem is, in this model, 'infinity' isn't a point, but a spherical surface.

If you've taken lower level undergraduate E&M, you know that electrostatic fields in source-free regions obey Laplace's equation, i.e.

div grad phi = 0

and that

E = grad phi,

where phi is the (scalar) electric potential. Alternatively, phi is the line integral of E.

By a vector identity, curl grad phi is identically zero. By Stokes' theorem, the potential difference along any closed path is the surface integral of the curl of the potential, so since curl phi = 0, the potential at any point is independent of how you got there. So the voltage is well defined everywhere.

The energy density of the field is proportional to the volume integral of |E|**2. (It's E**2 / 8pi in Gaussian units.)

In order for that to be finite, E has to go to zero at large distances faster than 1/r. (Actually it's asymptotically 1/r**2, and becomes purely radial very quickly--all tangential components die off as higher powers of r.)

Thus there's no voltage difference between points at large distances, so one point is as good as another.

Because of this, we adopt the simple convention that the potential at infinity is zero, allowing us to write

energy = 1/2 CV**2, where C is the self-capacitance.

We compute the self-capacitance by doing the volume integral to get the field energy and equating the two expressions. It's completely analogous to computing the self-inductance of a solenoid.

No giant spheres required, no 'other plate' need apply.

Cheers

Phil Hobbs

gradient, so since curl grad phi = 0, the

Coffee hadn't kicked in.

Cheers

Phil Hobbs

A couple of light-years away is probably good enough.

Need long test leads.

The moon will be far enough away. A conducting sphere at the moon orbit would change the Earth's C by less than 1%.

George H.

How would you make a decent e-field gradient sensor? Some sort of vibrating reed thing? Spinning ball with electrodes?

I bet there's e-fields everywhere. Insects? Plants? Doorknobs? Footprints on the floor? Bicycles? It would be cool if we could see e-fields.

use a field mill

Vibrating reed electrometers are commonly used to measure electrostatic voltages.

Cheers

Phil Hobbs

No (good) idea... An electric dipole in a gradient feels a force. I think an electric quadrupole in a E field gradient would feel a torque. (though it's a bit hard for me to 'see'.)

I never heard of vibrating reeds... This is a nice lecture/ chapter.

George H.

Sounds interesting. The flux from one capacitor, flows through another?

charging one capacitor puts a voltage on the other capacitor according to the ratio of the piezo transformer. but it's piezoelectric, it's not really a capacitance effect.

A dielectric fiber will polarize and align with E-field.. From field lines (because of Laplace's law) you can map gradients in uncharged spaces

Input data will look ike this:

Can you calibrate that?

Hmm I think that's ~1.5 long haired blonde, about 1kV/m in MKS units. :^)

George H.