Abs Values in Electric Potential & Potential Difference?

[chocopenguin]

I'm a little bothered with the inconsistency in notation of electric potential (V) and potential difference (ΔV) because they're apparently used synonymously... but what really confuses me more is that I've seen absolute value around ΔV sometimes. It may not matter (theoretically or mathematically) but I'd appreciate if somebody could clear it up for me!

For the following three equations, does it matter if the absolute value of ΔV is taken? Also, why?
$\Delta V=-\int_{a}^{b}\vec{E}\cdot \mathrm{d}\vec{l}$

$C=\frac{Q}{\Delta V}$

$U_{C}=\frac{1}{2}Q\Delta V=\frac{1}{2}C\Delta V^{2}$

For the latter two, I thought that ΔV would have to be positive even without the absolute value since it deals with capacitance... I don't think negative capacitance is possible? But for the first one, it doesn't seem to make sense to take the absolute value of ΔV since the negative of the integral of the field is taken.

 

[rcgldr]

A simple parallel plate capacitor separated by air can have positive or negative voltage (C = Q/V, both Q and V can be positive or negative, so C remains positive). Link to wiki image:

http://en.wikipedia.org/wiki/File:Plattenkondensator_hg.jpg

 

[Philip Wood]

ΔV or V ?
In all cases you've quoted, ΔV is a more logically correct choice than V, because where you take the zero of potential is irrelevant. But in the last two cases, as in much work with electric circuits, people tend to use just V. Apart from avoiding clumsiness, this frees up ΔV to use in a different sense, for example in the contexts of p.d.changing with time.

ΔV or |ΔV| ?
rcgldr has, I think, hit the nail on the head for C = Q/ΔV (or C = Q/V !). You may use the signed value of $\Delta$V, as long as you mean by$\Delta$V the change in potential going from the plate A to plate B, and by Q you mean the charge on plate B, for an arbitrary choice of which plate you're calling 'A' and which, 'B'. But C = |Q|/|$\Delta$V| would be equally good.

For your energy formula, it clearly doesn't matter whether you regard $\Delta$V as signed or not, since you're squaring it.