Mosfets with capacitor on drain?
This is a somewhat artificial problem, because you've shown no mechanism by which the capacitor is initially discharged so as to have no voltage across it, so it makes it difficult to intuitively understand the problem. But I'll imagine there is also an ideal switch across the capacitor that is closed for t &lt; 0 and opens at t = 0. 

After t = 0, the supply VS plays no further role in the behavior of the circuit on the other side of the capacitor, and we can ignore it. 

When the FET turns on (I'll assume you mean that VG is high enough to put the MOSFET in saturation), you can treat its channel as a very low value resistor. 

That means you will have a very low voltage at the drain, given by the resistor divider equation:

\$ V_D = V_1 \frac{R_{FET}}{R + R_{FET}}\$

Where I've arbitrarily named the node at the FET's drain as "D" and the node between the capacitor and the resitor as "1".

V1 will decay exponentially toward ground as usual for an RC circuit, and VD will decay proportionally.