The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy of the capacitor at the moment when the capacitor is half filled with the dielectric.
what i did is multiply by 1/2 to the equation with full with dielectric but i kept getting
-Your answer gives only the energy stored in the half of the capacitor that is filled with the dielectric. What is the energy stored in the other half?