1. The problem statement, all variables and given/known data A 24 V battery is used to charge two identical capacitors CL = CR = 6 µF, in parallel so Q(CL) = Q(CR) = 0.000144 C the battery is removed, and the energy stored in the system is .003456 J Next a dielectric with k = 1.4 is inserted between the plates of CR, fully filling the available space. (The dielectric carries zero net charge.) Calculate the new values of the charges on each capacitor. 2. Relevant equations C=Q/V (capacitance = charge/electric potential) C_new = C*k (capacitance = capacitance in a vacuum times K) capacitors in parallel add (C = c1 + c2) capacitors in series add their inverses (C =(1/C1 + 1/C2)^-1) 3. The attempt at a solution the capacitance of the system is (1/8.4 + 1/6)^-1 = 3.5 uF since the charge doesn't change, it's still 0.000144*2 C voltage over either capacitor is the same V*CL + V*CR = Q things break down from here. any pointers? thanks!