RC circuit with two voltage sources and finding I(t)
Checking the voltage loops you have: 

Loop 1 

$$2V - I_{R2}R2-I_{R1}R1=0$$

Loop 2 

$$1V - V_{cap} + I_{R1}R1=0$$

At t=0, there's no current flowing into the capacitor because it is already charged, meaning that \$I_{R2}=I_{R1}\$. Rewriting the first loop equation and solving for the current gives:

$$I_{R1}=\dfrac{2V}{R_1+R_2} = 500\mu A$$

Replacing the current in the second equation and solving for the capacitor voltage yields

$$V_{cap} = 1V + 500\mu A \cdot 1k\Omega=1.5V$$

So basically your capacitor will discharge from 1.5V to 1V. The discharging curve can be computed using:

$$V_{CAP, t}=V_{CAP,t=0}e^{\dfrac{-t}{1k\Omega \cdot 500pF}}$$

The rest you can figure out.

Here is a small simulation just to double check it.