Prior to t=0, switch is in position A (for long time).
Afterward, switch is in position B.

We're using \(v(t)=v(\infty)+[v(0)-v(\infty)]e^{\frac{-t}{\tau}}\) where \(\tau=RC\).

My problem is in the calculation of the thévenin resistance from the perspective of the capacitor.

My answer: \(R=R_\mathrm{th}=(15+10)\parallel 100=20\,\mathrm{k\Omega}\)
Homework answer: \(R=100\,\mathrm{k\Omega}\).

I was told that it has to do with the entire bottom of the circuit being at 0 V relative to the 100 V, so the left hand loop of the circuit is disabled. I'm not really satisfied with that answer though. I was hoping that someone could shed some light on this?

 

The switch disables the left hand side when in the "b" position. Follow the current flow.

 

The Thevenin equivalent from the capacitor's point-of-view is whatever the capacitor sees when going from one capacitor terminal until it reaches the other capacitor terminal. As Joe said, follow the current.