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

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.

• Is GH a short or is there an error in the diagram? Commented Jun 9, 2023 at 21:44

• The correct answer should be $\frac{V}{2}$ (2.5 volts) as the voltage across the capacitor between B and F. Commented Jun 9, 2023 at 19:33