An engineer working in an electronics lab connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 225 V. Assume a plate separation of d = 1.54 cm and a25.0 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.plate area of A =(a) Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.)beforepCafterQfpC%3D(b) Determine the capacitance (in F) and potential difference (in V) after immersion.Cf =AV, =%3D(c) Determine the change in energy (in nJ) of the capacitor.AU =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 225 V potential difference.Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.)beforepC%3DafterQfpCDetermine the capacitance (in F) and potential difference (in V) after immersion.Cf =FAV,V%3DDetermine the change in energy (in nJ) of the capacitor.AU =nJNeed Help?Read It

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