Re: Why does Capacitance decrease with "r" but F decreases with "r^2"?


Sue... wrote:
guskz@xxxxxxxxxxx wrote:
Sue... wrote:
guskz@xxxxxxxxxxx wrote:
Sue... wrote:
guskz@xxxxxxxxxxx wrote:
Wheter the capacitor's dielectric is space itself or another material,
why does Capacitance decrease with "r" but F decreases with "r^2"?


It would seem logical that if the attractive force between the charges
SUBSTANTIALLY decreases by r^2 then it doesn't seem logical that the
capacity to retain these charges(capacitance) in a capacitor decrease
only by r (even if space is the dielectric)?


F= kQq/r^2 (and capacitor's energy field: E = KQq/r^2) where as
Capacitance = Area/ (k * r)

<< These formulae are valid for any type of capacitor,
since the arguments we used to derive them do not
depend on any special property of parallel plate capacitors.
Where is the energy in a parallel plate capacitor actually
stored? Well, if we think about it, the only place it could be
stored is in the electric field generated between the plates.
This insight allows us to calculate the energy (or, rather,
the energy density) of an electric field.
Consider a vacuum-filled parallel plate capacitor ... >>
http://farside.ph.utexas.edu/teaching/302l/lectures/node34.html

Sue...

It doesn't explain the question? what it does at the end is w=cv^2/2 to
give a distance "r" instead of "r^2".

It doesn't explain why c (capacitance) and v( potential difference) are
only affected by distance "r" instead of "r^2" like F and the Energy
field?

Actually it says the Electric field = k Q/A where as other locations
E = kQq/r^2?

Spherical capacitors and granite slabs halfway to the earth's
centre are both pretty rare items but see if that visual and
the later exchange with Thomas Smid, will iron out where
your question is really just a matter of the geometry you
are assuming.

diatance 1
area 2
volume 3


How come the force between charges is AREA since Q & q are linearly
distant (= line) from each other where as the potential difference
force (V) is a line of force instead of an AREA of force?

The force is newtons. LOL
See the red interaction regions here:
http://newton.umsl.edu/~philf/trich_ex.gif
The stubby horizontal 'weiner' represents a volume.
When you slice the weiner in half you get a cross sectional
'area' that quantifies the interaction of the two charges.



Even if you had only one Q & q charge in total (instead of a plurality
of charges) the potential difference force (V) across Space (as a
dielectric) is a line of force (meaning your "distance 1" above)?

We speak of the force as acting along a *line of motion*.
But the charge interaction occurs in a 'volume' of space.
Slicing the volume quantifies the interaction as an 'area'.
where we might count imaginary flux lines whose quantity
would diminish by 1/r^2. ( 2 is exponenet for area Eh? )

You remember some visuals where the flux lines are ~countable~.

O +
(())
O -

Move the charges apart in free space:

\ /
O +

()

O -
/ \

Two of the flux lines found other partners out
in the 'sea of charges' that represents the ~vacuum~.

Image from:
http://newton.umsl.edu/~philf/triplet.html
Easier to visualize the volumes of space that
are operative, than is illustrated here:
http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/dipole.html#c2

As the advertisment says, it is easier to read in color.

Sue...

Ok I got mixed-up with forces and energy (sometimes called a force
itslef: electromotive force EMF or volts or force per unit charge).

The electrical potential = V = Fd (gives 1/r^1) where as the
electrical field = F/q = (gives 1/r^2)= V/r

So an Electrical field is a force field and not an energy field.

Still strange that Energy and Power Capacity (capacitance) only
decrease by a thousanth (1/r^1) when the total force that generates
them decreases by a millionth (1/r^2)



Your expected exponent is off by one so your
mental image condsiders too few or too many
dimensions somewhere.

Sue...
BTW the slab is weighless at the earths centre.

.

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