Homework Help is Here – Start Your Trial Now!
Science
Advanced Physics
A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 275 V. Assume a plate separation of d = 1.33 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, = v pC 457.47 after 457.47 V pC (b) Determine the capacitance (in F) and potential difference (in V) after immersion. C, = 1.33083e- v F AV, = 3.43 (e) Determine the change in energy (in nJ) of the capacitor. AU = (6.303e-8 × 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 275 V. Assume a plate separation of d = 1.33 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, = v pC 457.47 after 457.47 V pC (b) Determine the capacitance (in F) and potential difference (in V) after immersion. C, = 1.33083e- v F AV, = 3.43 (e) Determine the change in energy (in nJ) of the capacitor. AU = (6.303e-8 × Write expressions for the initial and final energies, and take the difference. nJ

icon
Related questions
Question
100%
A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 275 V. Assume a plate separation of d = 1.33 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, = 457.47 v pC after Q, = 457.47 v pC (b) Determine the capacitance (in F) and potential difference (in V) after immersion. C, = 1.33083e- Av, - 3.43 (c) Determine the change in energy (in nJ) of the capacitor. AU = 6.303e-8 Write expressions 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 275 V potential difference. Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before Q, 457.47 pC after Q, = 36586 V pC Determine the capacitance (in F) and potential difference (in V) after immersion. C; = 133.08 Will the capacitance be any different from what you found in part (b)? Note capacitance depends on the geometry and the dielectric constant, not the voltage. F AV, = 257 Re-read the problem statement. Note the source of potential difference remains attached to the capacitor in this case. V Determine the change in energy (in nJ) of the capacitor. AU = 6198.41 Write expressions for the initial and final energies, and take the difference. Note unlike part (c), the electric potential difference remains the same. 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 275 V. Assume a plate separation of d = 1.33 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, = 457.47 v pC after Q, = 457.47 v pC (b) Determine the capacitance (in F) and potential difference (in V) after immersion. C, = 1.33083e- Av, - 3.43 (c) Determine the change in energy (in nJ) of the capacitor. AU = 6.303e-8 Write expressions 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 275 V potential difference. Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before Q, 457.47 pC after Q, = 36586 V pC Determine the capacitance (in F) and potential difference (in V) after immersion. C; = 133.08 Will the capacitance be any different from what you found in part (b)? Note capacitance depends on the geometry and the dielectric constant, not the voltage. F AV, = 257 Re-read the problem statement. Note the source of potential difference remains attached to the capacitor in this case. V Determine the change in energy (in nJ) of the capacitor. AU = 6198.41 Write expressions for the initial and final energies, and take the difference. Note unlike part (c), the electric potential difference remains the same. nJ
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 5 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer