Circuits with Dielectric Problem 1. The problem statement, all variables and given/known data http://img412.imageshack.us/img412/2450/circuitgj6.jpg [Broken] The dielectric added has the value of 3.67. http://img504.imageshack.us/img504/5360/circuitquestionqw6.jpg [Broken] 2. Relevant equations C=Q/V C'=kC Series: 1/C=1/C1+1/C2... Parallel: C=C1+C2... 3. The attempt at a solution So, I can find the normal voltage and charge with the dielectric in, but once the battery is removed, I am lost as to how everything balances out. Let the capacitors be numbered: 1 for the top left, 2 for the top right, and 3 for the bottom. Let Cxy be the total capacitance of x and y. C12 = 1/(1/3.48 + 1/(3.67*8.8)) Removing the dielectric: C12' = 1/(1/3.48 + 1/8.8) Since V123 = V12, Q12 = C12' * V12 But this is not correct. Any tips on how everything balances out when you remove the dielectric? Thanks.