Charge in capacitor after switching voltage Say you have a capacitor connected to a voltage source (vs). After the capacitor has fully charged, you cut the source voltage in half to vs/2.
The voltage across the capacitor should change to match the source voltage (vc = vs/2). So what happens to the charge that was stored in the capacitor? Where did it discharge to?
My thought is that some equilibrium state will be reached, and the voltage at positive terminal will actually end up being slightly higher than vs/2. Is this right?
 A: 
Say you have a capacitor connected to a voltage source ($v_s$). After the capacitor has fully charged,...

If you have an ideal capacitor connected to an ideal voltage source, then there is no after.  By definition, the voltage at the terminals of the voltage source is constant, ($v_s$).  If the capacitor is connected across the voltage source, then the voltage across the capacitor's terminals must also be ($v_s$).

you cut the source voltage in half to $v_s/2$....The voltage across the capacitor should change to match the source voltage...So what happens to the charge that was stored in the capacitor? Where did it discharge to?

A current flowed from the positive terminal of the capacitor, through the voltage source, to the negative terminal.  Again, assuming ideal components, the impedance of the voltage source is $0\Omega$, it would be an infinite current for an infinitessimal amount of time.

My thought is that some equilibrium state will be reached, and the voltage at positive terminal will actually end up being slightly higher than vs/2. Is this right?

A practical circuit could be modelled by adding a non-zero resistance in series with the capacitor and the voltage source.  In that case, a finite current would begin to flow when you changed the voltage, and the current would exponentially decay.  After an arbitrarily long wait, the current would come arbitrarily close to zero, and the voltage across the capacitor would come arbitrarily close to $v_s/2$.
