How to find the voltage of a capacitor in a Op-amp + RC network?
Yes, everything is at zero volts up until t=0. What happens after that?



Somehow, you've managed to get the sign wrong for the final output voltage.

$$ \frac {0 - 6}{10} - \frac {6 - e}{10} = 0 $$

$$ \frac {0 - 6}{10} + \frac {e - 6}{10} = 0 $$

$$ 0 - 6 + e - 6 = 0$$

$$e = 12$$



Remember that the output of an ideal opamp is a voltage source. What does this mean in terms of the resistance "seen" by the capacitor?