How can a voltage across a capacitor be greater than the total voltage applied to the circuit? I was reading about complex impedance and AC circuits, and I just came across a sentence on a website stating that the voltage across a capacitor can be greater than the applied voltage. How can this be? I suppose that it may have something to do with phase differences, but I'm not entirely sure. 
 A: There are various ways this is possible, not including bookkeeping tricks like using RMS voltage in one place and peak in another.
With passive components, this can be achieved by resonance.  Consider the following simple circuit:

Energy can slosh back and forth between the inductor and capacitor forever at the resonant frequency.  If you were to break into this circuit and add even a small voltage in sync with the polarity of the current, the energy in the circuit would keep building up.  Both the maximum current thru the inductor and the maximum voltage on the capacitor would increase.  These would build indefinitely while you keep adding only a small voltage.
With active components there are more options, such as:

In fact, this is the basis of a circuit known as a boost converter, whose job is to make a higher voltage from a lower one.
Consider all currents zero at start, with the capacitor voltage V1.  When the switch closes, a constant voltage is applied to the inductor, so current increases linearly thru the inductor.  When the switch opens, this current must continue to flow instantaneously, which is thru the diode and the capacitor.  This charges up the capacitor.  The voltage on the cap rises as the first quadrant of a sine with the original voltage added.  Eventually the backwards voltage across the inductor causes the inductor current to go to zero.  At that point, all the energy originally in the inductor when the switch closed has been transferred to the capacitor.  This process can be repeated to reach arbitrarily large voltages on the capacitor in theory.
A: In typical discussion of AC voltage, we refer to the rms voltage.  So when you have a $120 V_{AC}$ line, the peak voltage is in fact $120\sqrt 2 \approx 170 V$  If you put a capacitor across the line, it will have an instantaneous voltage of $170 V$ at the peak.  If you drive a resonant circuit, the peak voltage can be even higher.
A: 
How can a voltage across a capacitor be greater than the total voltage applied to the circuit?

There are circuits with capacitors called Voltage multipliers.
One example is the Villard circuit (see the picture).

Villard circuit
A: A capacitor on a PSC induction motor which is wired in series with the start winding (and always in the circuit when running) will read higher than the applied voltage. This is due to the fact that although the cap is wired in series with the Start winding, it is also electrically connected across the Start and Run winding. The Start winding is up to 90 degrees out of phase with the run winding so the voltage read across the cap is an instantaneous reading of the two (additive) Voltages. You can test this for yourself with a simple Voltage measurement across the cap of a running PSC motor (Note that the cap may say "Run" on the cap - this is because it is always running in the ciruit. It is still wired in-series with the Start winding. If it also has a Start cap or caps, those are only in the circuit for starting and are removed by a PT xfmr or cent switch). My 5 ton a/c compressor motor has 337 Volts across the cap with a 240V applied voltage. I also checked a 2 hp in-ground well pump that had 286V with a 242 applied voltage. The total Volts will depend on how many degrees out of phase the two windings are. These cap voltages line up with the formulae for determining the Farad capacity of the cap:
Start winding amps X 2652/Volts across cap = MFD (of cap).
I've seen some suggestions that this higher voltage in these motor caps is due to CEMF. To my knowledge, CEMF is never higher than the applied voltage in a motor.
