1. The problem statement, all variables and given/known data E& M Purcell, 3.11 A 100-pF capacitor is charged to 100 volts. After the charging battery is disconnected, the capacitor is connected in parallel to another capacitor. If the final voltage is 30 volts, what is the capacitance of the second capacitor? How much energy was lost, and what happened to it? 2. Relevant equations Q = 1/4pi times integral of E dA or the integral of σ dA C = Q/phi not = a σ = E/4pi = δ1 - δ2 / s4pi capacitance of conductor: Q = C(δ1 - δ2) 3. The attempt at a solution Using the known equations and values and keeping in mind that the plate is charged in parallel, I know that the energy stored in a capacitor is U = 1/2 A/4pi(s) (Es)^2 or E^2/8pi * volume The voltage difference is 100-30 = 70 volts, so we start to plug in to find capacitance by denoting Q = 70, C = 70/phi. I don't really know how to find phi, or the energy afterwards, but I do know the energy equation. Any help would be greatly appreciated!