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 = 1.33 cm and a plate area of A = 25.0 cm?. 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 = 457.47 V pC after 457.47 v pC (b) Determine the capacitance (in F) and potential difference (in V) after immersion. C, = 1.33083e- F Av, = 3.43 (c) Determine the change in energy (in nJ) of the capacitor. AU = -62.11 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 275 V potential difference. Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before Q = 457.47 v pC after Q, = 36586 pC Determine the capacitance (in F) and potential difference (in V) after immersion. C, = 133.08 Will the capacitance be any different from what you found in part (b)? Note capacitance depends on the geometry and the dielectric constant, not the voltage. F AV, = 257 Re-read the problem statement. Note the source of potential difference remains attached to the capacitor in this case. V Determine the change in energy (in nJ) of the capacitor. AU = 6198.41 Write expressions for the initial and final energies, and take the difference. Note unlike part (c), the electric potential difference remains the same. nJ

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 = 1.33 cm and a plate area of A = 25.0 cm?. 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 = 457.47 V pC after 457.47 v pC (b) Determine the capacitance (in F) and potential difference (in V) after immersion. C, = 1.33083e- F Av, = 3.43 (c) Determine the change in energy (in nJ) of the capacitor. AU = -62.11 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 275 V potential difference. Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before Q = 457.47 v pC after Q, = 36586 pC Determine the capacitance (in F) and potential difference (in V) after immersion. C, = 133.08 Will the capacitance be any different from what you found in part (b)? Note capacitance depends on the geometry and the dielectric constant, not the voltage. F AV, = 257 Re-read the problem statement. Note the source of potential difference remains attached to the capacitor in this case. V Determine the change in energy (in nJ) of the capacitor. AU = 6198.41 Write expressions for the initial and final energies, and take the difference. Note unlike part (c), the electric potential difference remains the same. nJ