A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 275 V. Assume a plate separation of d = 1.53 cm and a plate area of A = 25.0 cm². When the battery is removed, the capacitor is plunged into a container of distilled water. Assume distilled water is an insulator with a dielectric constant of 80.0. (a) Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before Q after = Qf = PC рс (b) Determine the capacitance (in F) and potential difference (in V) after immersion. CF = AV₂ = F (c) Determine the change in energy (in n]) of the capacitor. AU = n] (d) What If? Repeat parts (a) through (c) of the problem in the case that the capacitor is immersed in distilled water while still connected to the 275 V potential difference. Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before Q after = Qf = PC PC Determine the capacitance (in F) and potential difference (in V) after immersion. = C₁ = AV₂ = F V Determine the change in energy (in n]) of the capacitor. AU = nJ