How do capacitors affect Operational Amplifiers?
If V2 is DC, we do not have any AC source on the circuit. Neglecting the transients, when you only have DC sources the capacitors act as open circuits, while the inductors are closed circuits.

Think it like that: you connect your DC sources, turn them on, caps charge up, current ramps up in the coils, and at a certain point steady state is reached: dV/dt = 0 means i = 0 in caps, and conversely V = 0 in inductors -> open circuits, closed circuits.

Now, if you remove C1 and C2 from your schematic, you can remove also R1 because one end is not connected. No current can flow in R2 because no current can flow in the inverting input of the amplifier. No current in a resistor means that the voltage across it is 0, so you can see it as a short circuit -> you have your buffer, and as you know VA = V2.

Apply the same reasoning of R2 to RL and you also get VB = 0.