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A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 255 V. Assume a plate separation of d = 1.55 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 = 364 ✓ PC after Qf=29133 x The charge will be the same before and after the capacitor is submerged in water. Apply the definition of a parallel-plate capacitor, and the relationship among C, Q, and AV. Don't forget to convert from cm to m as needed. pC (b) Determine the capacitance (in F) and potential difference (in V) after immersion. C₁ = 1.142e-10 F AV = 255 X This is the potential difference before the capacitor was submerged. When it is in the distilled water, the water acts as a dielectric. You need to take this into account when calculating AV. V (c) Determine the change in energy (in nJ) of the capacitor. ÂU = | 3668 Write expressions for the initial and final energies, and take the difference. nJ

A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 255 V. Assume a plate separation of d = 1.55 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 = 364 ✓ PC after Qf=29133 x The charge will be the same before and after the capacitor is submerged in water. Apply the definition of a parallel-plate capacitor, and the relationship among C, Q, and AV. Don't forget to convert from cm to m as needed. pC (b) Determine the capacitance (in F) and potential difference (in V) after immersion. C₁ = 1.142e-10 F AV = 255 X This is the potential difference before the capacitor was submerged. When it is in the distilled water, the water acts as a dielectric. You need to take this into account when calculating AV. V (c) Determine the change in energy (in nJ) of the capacitor. ÂU = | 3668 Write expressions for the initial and final energies, and take the difference. nJ

Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter20: Electric Potential And Capacitance
Section: Chapter Questions
Problem 64P
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A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 255 V. Assume a plate separation of d = 1.55 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 = 364 PC after Qf = 29133 The charge will be the same before and after the capacitor is submerged in water. Apply the definition of a parallel-plate capacitor, and the relationship among C, Q, and AV. Don't forget to convert from cm to m as needed. pC (b) Determine the capacitance (in F) and potential difference (in V) after immersion. Cf= 1.142e-10 F X AV f = 255 This is the potential difference before the capacitor was submerged. When it is in the distilled water, the water acts as a dielectric. You need to take this into account when calculating AV. V (c) Determine the change in energy (in nJ) of the capacitor. AU = |3668 X Write expressions for the initial and final energies, and take the difference. 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 255 V potential difference. Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before Q₁ pC after Qf = pC Determine the capacitance (in F) and potential difference (in V) after immersion. Cf = F AV f = V Determine the change in energy (in nJ) of the capacitor. AU = nJ
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 255 V. Assume a plate separation of d = 1.55 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 = 364 PC after Qf = 29133 The charge will be the same before and after the capacitor is submerged in water. Apply the definition of a parallel-plate capacitor, and the relationship among C, Q, and AV. Don't forget to convert from cm to m as needed. pC (b) Determine the capacitance (in F) and potential difference (in V) after immersion. Cf= 1.142e-10 F X AV f = 255 This is the potential difference before the capacitor was submerged. When it is in the distilled water, the water acts as a dielectric. You need to take this into account when calculating AV. V (c) Determine the change in energy (in nJ) of the capacitor. AU = |3668 X Write expressions for the initial and final energies, and take the difference. 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 255 V potential difference. Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before Q₁ pC after Qf = pC Determine the capacitance (in F) and potential difference (in V) after immersion. Cf = F AV f = V Determine the change in energy (in nJ) of the capacitor. AU = nJ
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