In the above circuit, is it correct to assume that since capacitors in DC circuits act as an open, the diode in the circuit cannot be forward or reverse bias since there is no current flowing through it, therefore it is as if it isn't even there?

 

Depends on the initial state of the capacitor when the circuit is first energized by the source.

 

Assume it's completely depleted and t = 0. Cap should act as an open since by definition I = C*dV/dt and since the voltage source is not changing over time, dV/dt = 0 which means the current through the capacitor is 0. This further goes to show me that since the diode is not getting any current (due to being in series with the cap) forward or reverse bias is not possible since the current is 0.

I'm just not sure if my thought process is correct.

 

So how do you explain the well established phenomenon of charging a capacitor from a DC supply? For example capacitors are routinely used in DC powered circuits.

Even if the diode were reversed one would expect the capacitor to eventually charge, with the real diode reverse saturation current being the prime determinant of the rate of charge. Measuring the voltage would present some problems.

An issue with the idealized schematic you post is that it is conceptually difficult to meaningfully apply the relationship I=C*dv/dt. Imagine the circuit being energized at some time t=0. One would assume the diode is meant to be ideal and hence acts like a switch - conducting if forward biased. With nothing to limit the current and the capacitor initially uncharged the current flow would comprise an infinite impulse or delta / Dirac function. Consider the case of a capacitor charged from a DC supply via a current limiting resistor plus the (ideal) diode and then you have a more realistic perspective.

 

Assume it's completely depleted and t = 0. Cap should act as an open since by definition I = C*dV/dt and since the voltage source is not changing over time, dV/dt = 0 which means the current through the capacitor is 0. This further goes to show me that since the diode is not getting any current (due to being in series with the cap) forward or reverse bias is not possible since the current is 0.

I'm just not sure if my thought process is correct.

dv/dt is not referring to a change in voltage source, but the change in voltage drop across the capacitor with respect to the change in time.