7. [4/10 Points] PREVIOUS ANSWERS SERPSE1O 25.5.0P.018. ASK YOUR TEACHER PRACTICE ANOTHER A student working in the physics laboratory connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 265 V. Assume a plate separation ofd = 1.36 cm and a plate area ofA = 25.0 cmz. 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 aer the capacitor is submerged. (Enter the magnitudes.) before Q,- = 431.95 ,/ pC after Q,r = 431.95 v pC (b) Determine the capacitance (in F) and potential difference (in V) after immersion. Cf = 130.4e-12 'I F AVf = 3.3125 V V (c) Determine the change in energy (in nJ) of the capacitor. AU = 4.647 x Write expressions for the initial and final energies, and take the difference. nJ (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 265 V potential difference. Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before Q,- = pC after Q, = pC Determine the capacitance (in F) and potential difference (in V) after immersion. C; = C F AVf = V Determine the change in energy (in nJ) of the capacitor. Mn: Need Help?