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

*From*: "Sue..." <suzysewnshow@xxxxxxxxxxxx>*Date*: 19 Jun 2006 02:16:32 -0700

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...

.

**Follow-Ups**:**Re: Why does Capacitance decrease with "r" but F decreases with "r^2"?***From:*Igor

**Re: Why does Capacitance decrease with "r" but F decreases with "r^2"?***From:*guskz@xxxxxxxxxxx

**Re: Why does Capacitance decrease with "r" but F decreases with "r^2"?***From:*Thomas Smid

**References**:**Why does Capacitance decrease with "r" but F decreases with "r^2"?***From:*guskz@xxxxxxxxxxx

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