A parallel-plate capacitor has plates with an area of 445 cm2 and an air-filled gap between the plates that is 1.51 mm thick. The capacitor is charged by a battery to 575 V and then is disconnected from the battery. There is 4.311536113 x 10 ^-5 J of energy stored in the capacitor. The separation between the plates is now increased to 3.02 mm. How much energy is stored in the capacitor now?
I know that since the capacitor is disconnected from the battery, V can change with C while Q, being trapped in the disconnected
capacitor, is constant. I calculated Q for the original
capacitor, which was about 1.5 x 10^-7 C. When the spacing, d, between the plates is increased, C changes, and V changes with it, so I know that I can't use the equation U=(1/2)CV^2 (with V=575 V) to find the energy, but I'm not sure exactly what I can use.