How do capacitors discharge? I'm having some trouble understanding how capacitors work. I can visualize how a capacitor would charge up and how current "flows" through it.
But I don't really understand the concept of the discharge. I gather that you simply get rid of the accumulated charge on the plates, and it happens relatively fast. But how? Do you get a spark?
And on top of that, in a simple circuit with the battery, a capacitor and a light bulb connected in series, how does the discharge of the capacitor make a bright flash of the bulb (the typical example of a use of a capacitor).
Thanks a lot!
 A: Okay, so capacitors allow current to flow, but not electrons to flow, between the terminals. As a consequence, when they are operational, a density of electrons builds up on one side and a density of missing-electrons builds up on the other side, until the charge on either plate is given by $|Q| = C |V|$ where $C$ is the capacitance and $V$ is the voltage across the junction.
Suppose we have a loop, as you say:
  +----[Battery, V1]>----+
  |                      |
===== C                  |
  |                      |
  +----[Light Bulb, R]---+

This is a typical RC-circuit; if the voltage across the capacitor starts out as $V_0$ then the voltage as a function of time is:
$ V(t) ~=~ V_1 ~+~ (V_0 - V_1) e^{-t/RC} $
This means that most of the "action" happens over the time scale $R C$ where $R$ is the resistance of the light bulb and $C$ is again the capacitance.
So let's start off with $V_0 = 0$, $V_1 = V^*$ for some special voltage $V^*$. The light bulb will shine some light as the capacitor charges, until the capacitor's electrons are built up enough to "push against the battery"; then $V = V^*$ as $t \rightarrow \infty$.
Once we have reached this steady state, let's disconnect the battery completely and insert a short where the battery was. Then it's the same story with $V_0 = V^*$ and $V_1 = 0$. Current flows backwards across the light bulb, which would also light it up. 
You do not get a spark unless the capacitor breaks. Very important. The electrons do not jump the gap. They return to the other side by going back the way they came.
A: 
I gather that you simply get rid of the accumulated charge on the
  plates, and it happens relatively fast.

If a capacitor is discharging, current exits the more positive terminal rather than entering.  That's really all there is to it.
When current enters the more positive terminal, power is delivered to the capacitor and, thus, the stored energy increases.
When current exits the more positive terminal, power is supplied by the capacitor and, thus, the stored energy decreases.
This does not need to occur quickly.  If you connect a resistor with resistance $R$ across a charged (and otherwise unconnected) capacitor with capacitance $C$, the time required to discharge the capacitor to less than 1% of its initial charge is about $t = 5RC$.
For example, let $R = 1\mathrm{k\Omega}$ and $C = 1\mathrm{\mu F}$.  The time required to discharge the capacitor is
$$t_D = 5\cdot 1\mathrm{k\Omega}\cdot1\mathrm{\mu F} = 5\mathrm{ms}$$
