How do I find the charge in the capacitor between B and F in this circuit?

Question

Where each resistor has a resistance of \$1000K \Omega\$ and each capacitor having a capacitance of \$4 \mu F \$

This is what I did:

Resistors AB, BC, CG, GF, FE, EH and HD are all in series across AD, so an equivalent resistance of \$7000\$ across AD.

Capacitors BF and AE are in parallel, and their equivalent would be \$ 8 \mu F \$, which would be in series with capacitor CD, from which I concluded that the potential drop across BF is half that of DH (In a capacitor, potential is inversely related to capacitance) which would imply that the potential drop across BF is \$ \frac{5}{3} V \$, which would mean that the potential drop across BF is \$ \frac{20}{3}V\$, however, my book says this is wrong.

Answer 1

If you are after the steady state voltage then...

Then redraw and calculate the open circuit voltage across B and F.

That will be your answer. It looks fairly easy now.