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e 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, = 353 Q = 353 v pC v pC after b) Determine the capacitance (in F) and potential difference (in V) after immersion. C; = 1.26e-10 AV, - 2.81 V c) Determine the change in energy (in n)) of the capacitor. AU = 0.124 x n)

e 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, = 353 Q = 353 v pC v pC after b) Determine the capacitance (in F) and potential difference (in V) after immersion. C; = 1.26e-10 AV, - 2.81 V c) Determine the change in energy (in n)) of the capacitor. AU = 0.124 x n)

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P5555

Please solve Part(c).

Note: 39.3 & 0.124 are proven to be wrong answers for part(c)

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 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 Q; 353 V pC after Qf = 353 pC (b) Determine the capacitance (in F) and potential difference (in V) after immersion. Cf = 1.26e-10 F AV = 2.81 V (c) Determine the change in energy (in nJ) of the capacitor. AU = 0.124 X 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: = 353 pC after Qf = 28242 pC Determine the capacitance (in F) and potential difference (in V) after immersion. Cf = 1.26e-10 F AVE = 225 Determine the change in energy (in n)) of the capacitor. AU = 3137 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 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 Q; 353 V pC after Qf = 353 pC (b) Determine the capacitance (in F) and potential difference (in V) after immersion. Cf = 1.26e-10 F AV = 2.81 V (c) Determine the change in energy (in nJ) of the capacitor. AU = 0.124 X 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: = 353 pC after Qf = 28242 pC Determine the capacitance (in F) and potential difference (in V) after immersion. Cf = 1.26e-10 F AVE = 225 Determine the change in energy (in n)) of the capacitor. AU = 3137 nJ
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