student working in the physics laboratory connécts a parallel-plate capacitor to a battery, so that the potential difference between the plates is 225 V. Assume a plate separation of d= 1.41 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, = pC after Q = pC (b) Determine the capacitance (in F) and potential difference (in V) after immersion. C, = AV; = V (c) Determine the change in energy (in n) 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.) before Q, = pC after Q; = pC Determine the capacitance (in F) and potential difference (in V) after immersion. C, = AV, = V Determine the change in energy (in n) of the capacitor. A =| n)

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Chapter1: Units, Trigonometry. And Vectors
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A student working in the physics laboratory 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 = 1.41 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
pC
after
Q =
pC
(b) Determine the capacitance (in F) and potential difference (in V) after immersion.
Cf =
F
AV =
V
(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.)
before
Q; =
pC
after
Qf
pC
Determine the capacitance (in F) and potential difference (in V) after immersion.
Cf =
AV =
V
Determine the change in energy (in nJ) of the capacitor.
Δυ-
nJ
Transcribed Image Text:A student working in the physics laboratory 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 = 1.41 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 pC after Q = pC (b) Determine the capacitance (in F) and potential difference (in V) after immersion. Cf = F AV = V (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.) before Q; = pC after Qf pC Determine the capacitance (in F) and potential difference (in V) after immersion. Cf = AV = V Determine the change in energy (in nJ) of the capacitor. Δυ- nJ
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