) 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, = 357.6 Q, - 28608 pC v pC after Determine the capacitance (in F) and potential difference (in V) after immersion. C; = 00000000010 , F AV; = 268.3 Determine the change in energy (in n) of the capacitor. AU = 585 Write expressions for the initial and final energies, and take the difference. Note unlike part (c), the electric potential difference remains the same. nJ
) 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, = 357.6 Q, - 28608 pC v pC after Determine the capacitance (in F) and potential difference (in V) after immersion. C; = 00000000010 , F AV; = 268.3 Determine the change in energy (in n) of the capacitor. AU = 585 Write expressions for the initial and final energies, and take the difference. Note unlike part (c), the electric potential difference remains the same. nJ
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Could I get help with the last section, d? I just can't figure out the energy change. Thanks!
Transcribed Image Text:A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 265 V. Assume a plate separation of d = 1.64 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
357.6
Q;
357.6
pC
after
Qf
pC
(b) Determine the capacitance (in F) and potential difference (in V) after immersion.
.00000000010
Cf
= 3.312
F
AVF
V
(c) Determine the change in energy (in nJ) of the capacitor.
AU = -46.7
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
357.6
pC
after
Qf
= 28608
pC
Determine the capacitance (in F) and potential difference (in V) after immersion.
Cf =
.00000000010
F
AVE = 268.3
V
Determine the change in energy (in nJ) of the capacitor.
AU = .585
Write expressions for the initial and final energies, and take the difference. Note unlike part (c), the electric potential difference remains the same. nJ
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