Naked Science Forum

Non Life Sciences => Technology => Topic started by: jaiii on 28/02/2011 08:52:15

Title: Casimir plate
Post by: jaiii on 28/02/2011 08:52:15

Hello.
It is possible to create a rectangle with a conductive plate which has external dimensions of 100 m * 50 m * 2 m and filled with conductive surfaces inside 100m * 2 m spaced 5um 5*10^6 m?
Or conductive tubing diameter 5um 5*10^-6 m?

Thank you

Title: Casimir plate
Post by: Jolly- Joliver on 30/03/2011 01:01:12

Quote from: jaiii on 28/02/2011 08:52:15

Hello.
It is possible to create a rectangle with a conductive plate which has external dimensions of 100 m * 50 m * 2 m and filled with conductive surfaces inside 100m * 2 m spaced 5um 5*10^6 m?
Or conductive tubing diameter 5um 5*10^-6 m?

Thank you

So you want a 100m by 50m box 2 metres thick. Hollow inside and filled with conductive tubes set at 5micrometres apart(5*10^-6) I take it that is the tube diamensions.

I don't think there is enought info there really need more specs. It sound like you seek to run electricity through it and achieve something within the tube system.

But I'm sure it can be build.

Title: Casimir plate
Post by: jaiii on 30/03/2011 09:14:34

It is for Casimir event .Thank.

Title: Casimir plate
Post by: imatfaal on 30/03/2011 12:28:33

Jaiii the force developed by casimir effect is related to the inverse 4th power of the seperation of the plates.  That means that at a micrometre seperation you are unlikely to get a meaningful force.

On a very boring telephone conversation - so here goes with the sums

F = -1/240(A.hbar.c.π^2.s^-4)

F = force
A = Area
c = speed of light
hbar = reduced plancks constant
π = pi
s = seperation

in same order as above eq
F = -1/240 (100*50).(1*10^-34).(3*10^8).(3.14)^2.(5*10^-6)^-4

F = -1/240 (5*10^3).(1*10^-34).(3*10^8).(10^2).(6.25*10^22)

F = -1/240 (15/6)*10^(3-34+8+2+22)

F = -1/240 (2.5*10^1)

F = -.01

(to explain long-winded route - all done by hand cos I am a masochist)

let's check units via dimensional analysis

F = -1/240(A.hbar.c.π^2.s^-4)
  =  m^2 . js . ms-1. m-4
  =  jm-1

Joules per metre is the Newton.  So in effect your football field apparatus which would cost multiple millions - (and remember it's gotta be in a hard vacuum)  will generate a force of around one hundredth of a newton.

Title: Casimir plate
Post by: jaiii on 30/03/2011 14:46:54

I am veri sorry gap between plate is 10^-9 [nm]
But I thank for your help . I try this counting with this numbers.

I say 10^-6 becouse I dont veri thats 10^-9 is possible.

Thank veri much and if everyone were like you too.

BTW Can i use other number as is ?

Title: Casimir plate
Post by: imatfaal on 30/03/2011 15:32:23

At nanometre scales you start getting decent forces (12 orders of magnitude higher than micrometre) - but and its a very big but - you cannot manufacture a pair of plates that are 100m long to an accuracy of 10nm! 

To get an idea of scale 10nm is probably less than a hundred atoms - and that's your tolerance over an area the size of a football pitch

Title: Casimir plate
Post by: jaiii on 30/03/2011 16:01:46

Therefore, the long time I got no response.
So I changed the configuration.
The total length = 100 m
Height of the plate 10 m
Plate width 5 m
There is several million plate gap with 0-20 nm.
The plates are hung in a frame length 100m
and moves with the help of magnetic field.

But even so I still thank you very much again for answers and for computation.
And even the boring interview.

Hello.