Talk:Capacitance

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Wrong expressions or definitions for two-sphere case?

Taking the first two terms of the Taylor series version leads to an expression that agrees with what one calculates assuming that the charges are uniformly distributed on each sphere, but only if the value d is defined as the gap between the spheres' surfaces. As written, it looks like d is trying to be the distance between the centers of the spheres (given the d>2a constraint). The definition of d may therefore be wrong here. However, I cannot access either cited source for the expression(s). Perhaps someone who can access these can take a look.

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Yes, d is the center-to-center separation.
No, the charges will not be uniformly distributed.

Constant314 (talk) 03:30, 4 April 2025 (UTC)[reply]

DISC and thin WIRE CAPACITANCE

Propound : for Jackson and Cavendish the ratio between the sphere capacitance and the disc capacitance is 4pi/8... But ,I think that this ratio have to be the square of something because the charge disc to sphere depend of a double integral of sigma/r ds , double because the transition of 2 Dimensions (disc) to 3 Dimensions (sphere)...and the ratio I propose is the square of 4/pi where 4 is the ratio of sphere and disc surfaces and 1/pi is the ratio of sphere to disc density of charges... So pi3/4 of disc multiplied by square of 4/pi is equal to 4pi of the sphere... By extension the thin wire capacitance is pi3/4 x epsilon zero x l/ln l/a... All of my humble propound is easy to measure... — Preceding unsigned comment added by ~2026-30426-50 (talk) 17:21, 21 May 2026 (UTC) edit by leveugel — Preceding unsigned comment added by Leveugel (talk • contribs) 17:24, 21 May 2026 (UTC)[reply]

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It cannot go into Wikipedia without a reliable source.

Constant314 (talk) 18:28, 21 May 2026 (UTC)[reply]

Capacitance relative progression

It seems that the progression between circular disc and square disc, cone ,sphere, cylinder and cube is in

(4/pi) to the power n....n is 0.5 for square disc, 1.5 for cone, 2 for sphere, 2.5 for cylinder and 3 for cube... Leveugel (talk) 17:12, 26 May 2026 (UTC)[reply]

So , by example , Disc capacitance = pi3 / 4 x epsilon0 x radius...Hemisphere = pi2 x E0 x r...Sphere = 4pi x E0 x r...and Cube = 16 x E0 x r ...Leveugel (talk) 08:59, 7 June 2026 (UTC)[reply]

As before, it cannot go into Wikipedia without a reliable source. Constant314 (talk) 17:15, 26 May 2026 (UTC)[reply]

Capacitance of Circular Loop

In the Article there is no Formula for this basic shape...

I don't find any Reference on the topic...

The basic formula I use is C= pi/4 x E0 x l divided by ln l/r

It correspond with the straight wire capacitance divided by pi2...

My only goal is to help to improve the Article in this specific way .Leveugel (talk) 11:37, 12 June 2026 (UTC)[reply]

See section section 1.4. Toroidal Surface of [1], and take the limit for

a\ll A

. catslash (talk) 16:56, 13 June 2026 (UTC)[reply]

Thank's a lot for your proposal...

It concern the static capacitance of a loop..

So I am still looking for the dynamic capacitance when there is a current circulation in the loop. Leveugel (talk) 08:48, 14 June 2026 (UTC)[reply]