(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 270 V potential difference. Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before Q₁ = | after pC PC Determine the capacitance (in F) and potential difference (in V) after immersion. Cf AV, = Determine the change in energy (in nj) of the capacitor. V
(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 270 V potential difference. Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before Q₁ = | after pC PC Determine the capacitance (in F) and potential difference (in V) after immersion. Cf AV, = Determine the change in energy (in nj) of the capacitor. V
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Could you please help me out with part What if (d) and so on?
Please write out the solution,that way I can follow along and understand the steps, if it's in the "typed out" format is confusing to follow the thought process. Thank you so much!!
![An engineer working in an electronics lab connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 270 V. Assume a plate separation of d = 1.61 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 Q₁
after Qf =
=
(b) Determine the capacitance (in F) and potential difference (in V) after immersion.
Cf =
AV f =
F
V
pC
pC
(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 270 V potential difference.
Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.)
before Q₁
pC
pC
after Qf =
Determine the capacitance (in F) and potential difference (in V) after immersion.
Cf
AV f =
F
V
Determine the change in energy (in nJ) of the capacitor.
AU = |
n]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F76aaab3c-5a4a-4ff8-814d-d2e8a1f30217%2F4e06f693-76f8-4681-ba87-03d46b969057%2Fg2w1d1_processed.png&w=3840&q=75)
Transcribed Image Text:An engineer working in an electronics lab connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 270 V. Assume a plate separation of d = 1.61 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 Q₁
after Qf =
=
(b) Determine the capacitance (in F) and potential difference (in V) after immersion.
Cf =
AV f =
F
V
pC
pC
(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 270 V potential difference.
Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.)
before Q₁
pC
pC
after Qf =
Determine the capacitance (in F) and potential difference (in V) after immersion.
Cf
AV f =
F
V
Determine the change in energy (in nJ) of the capacitor.
AU = |
n]
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