Talk about "current leading voltage" or "phase difference" only applies to AC analysis. In the more general case, one could say what a capacitor really does is [differentiate] voltage, according to:

$$i = C\frac{dv}{dt}$$

From this, you can derive all sorts of well-known things about capacitors. Such as, if you want a linearly changing voltage across a capacitor, you must apply a constant-current source to it. As an example, consider a 1 ampere current source connected to a 1 farad capacitor:

\require{cancel} \begin{align} 1A &= 1F \frac{dv}{dt} \\ 1A &= \frac{1 A \cdot s}{V} \frac{dv}{dt} \\ \frac{1\cancel{A}\cdot V}{1\cancel{A}\cdot s} &= \frac{dv}{dt} \\ \frac{1V}{s} &= \frac{dv}{dt} \end{align}

If you consider the case where the applied voltage is sinusoidal, then so too is the current:

\begin{align} i &= C\frac{dv}{dt} \\ i &= C\frac{d\sin(t)}{dt} \\ i &= C\cos(t) \end{align}

[because \$\cos\$ is the derivative of \$\sin\$].

You will also see if you [graph these functions], that \$\cos\$ (current) leads \$\sin\$ (voltage) by 90 degrees, as an electrical engineer would put it:

![plot of sin and cos]

: http://en.wikipedia.org/wiki/Differentiation_%28mathematics%29
: http://en.wikipedia.org/wiki/Differentiation_of_trigonometric_functions
: http://www.wolframalpha.com/input/?i=graph+sin+t%2C+cos+t
: http://i.stack.imgur.com/aRkD2.png