Find the charge in the plates before the glass is removed. Find the energy stored in the capacitor before the glass is removed.Find the charge in the plates after the glass is removed. Find the energy stored in the capacitor after the glass is removed.Find the potential difference after the glass is removed.

Required information A parallel plate capacitor of capacitance 5.79 UF has the space between the plates filled with a slab of glass with K = 3.00. The capacitor is charged by connecting it to a 1.40-V battery. After the capacitor is disconnected from the battery, the dielectric slab is removed. Glass (1) (2) (3) X = 1.40 V