Two 2.2-cm-diameter-disks spaced 2.0 mm apart form a parallel-platecapacitor. The electric field between the disks is 4.3×105 V/mFor the steps and strategies involved in solving a similar problem, youmay view a Video Tutor Solution.▼Part AWhat is the voltage across the capacitor?Express your answer with the appropriate units.V = 860 VSubmit✓ CorrectImportant: If you use this answer in later parts, use the full unrounded value in your calculations.Part BPrevious AnswersHow much charge is on each disk?Express your answers in coulombs separated by a comma.SubmitPart CIVE ΑΣΦxaXb+ √x vx xxPrevious Answers Request Answer1X191.921.45 10-⁹Your submission doesn't have the correct number of answers. Answers should be separated with a comma.No credit lost. Try again.www. ?X.10nсAn electron is launched from the negative plate. It strikes the positive plate at a speed of 2.0x107 m/s. What was the electron's speed as it left the negative plate?Express your answer with the appropriate units.

Answered: Image /qna-images/answer/3b3517f3-a1e9-4434-ac20-a9d438fecc7b.jpg

Two 2.2-cm-diameter-disks spaced 2.0 mm apart form a parallel-plate capacitor. The electric field between the disks is 4.3x105 V/m For the steps and strategies involved in solving a similar problem, you may view a Video Tutor Solution. What is the voltage across the capacitor? Express your answer with the appropriate units. V = 860 V Submit Previous Answers ✓ Correct Important: If you use this answer in later parts, use the full unrounded value in your calculations. Part B How much charge is on each disk? Express your answers in coulombs separated by a comma. 195] ΑΣΦ x² xb = √x VXxx xxx-10m ? Submit Previous Answers Request Answer BITC 91-92 1.4510-⁹ Your submission doesn't have the correct number of answers. Answers should be separated with a comma. No credit lost. Try again. с с

Two 2.2-cm-diameter-disks spaced 2.0 mm apart form a parallel-plate capacitor. The electric field between the disks is 4.3x105 V/m For the steps and strategies involved in solving a similar problem, you may view a Video Tutor Solution. What is the voltage across the capacitor? Express your answer with the appropriate units. V = 860 V Submit Previous Answers ✓ Correct Important: If you use this answer in later parts, use the full unrounded value in your calculations. Part B How much charge is on each disk? Express your answers in coulombs separated by a comma. 195] ΑΣΦ x² xb = √x VXxx xxx-10m ? Submit Previous Answers Request Answer BITC 91-92 1.4510-⁹ Your submission doesn't have the correct number of answers. Answers should be separated with a comma. No credit lost. Try again. с с

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter27: Capacitors And Batteries

Section: Chapter Questions

Problem 40PQ

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Similar questions

(a) How much charge is on each plate of a 4.00-F capacitor when it is connected to a 12.0-V battery? (b) If this same capacitor is connected to a 1.50-V battery, what charge is stored?
(a) Will the electric field strength between two parallel conducting plates exceed the breakdown strength of dry air, which is 3.00106 V/m, if the plates are separated by 2.00 mm and a potential difference of 5.010V is applied? (b) How close together can the plates be with this applied voltage?
At a certain distance from a charged particle, the magnitude of the electric field is 500 V/m and the electric potential is 3.00 kV. (a) What is the distance to the particle? (b) What is the magnitude of the charge?
In three regions of space, the electric potential is given by V(r)=0forrRV(r)=V04R2r2forRr2RV(r)=V0forr2R a. Plot V as a function of r. b. Find expressions for the electric field in all three regions. c. Plot E versus r in all three regions.
(a) Calculate the electric potential 0.250 cm from ail electron, (b) What is the electric potential difference between two points that are 0.250 cm and 0.750 cm from an electron? (c) How would the answers change if the electron were replaced with a proton?
When a 360-nF air capacitor is connected to a power supply, the energy stored in the capacitor is 18.5J . While the capacitor is connected to the power supply, a slab of dielectric is insetted that completely fills die space between the plates. This increases the stored energy by 23.2J . (a) What is the potential difference between the capacitor plates? (b) What is die dielectric constant of the slab?
(a) Will the electric field strength between two parallel conducting plates exceed the breakdown strength for air ( 3.0 106 V/m) if the plates are separated by 2.00 mm and a potential difference of 5.0 103 V is applied? (b) How close together can the plates be with this applied voltage?
A very large sheet of insulating material has had an excess of electrons placed on it to a surface charge density of 3.00nC/m2 . (a) As the distance from the sheet increases, does the potential increase or decrease? Can you explain why without any calculations? Does the location of your reference point matter? (b) What is the shape of the equipotential surfaces? (c) What is the spacing between surfaces that differ by 1.00 V?
Suppose that the capacitance of a variable capacitor can be manually changed from 100 to 800 pF by turning a dial connected to one set of plates by a shaft, from 0 to 180 . With the dial set at 180 (corresponding to C = 800 pF), the capacitor is connected to a 500-V source. After charging, the capacitor is disconnected from the source, and the dial is turned to 0°. (a) What is the charge on the capacitor? (b) What is the voltage across the capacitor when the dial is set to 0°?
(a) Find the electric potential difference Ve required to stop an electron (called a stopping potential) moving with an initial speed of 2.85 107 m/s. (b) Would a proton traveling at the same speed require a greater or lesser magnitude of electric potential difference? Explain. (c) Find a symbolic expression for the ratio of the proton stopping potential and the electron stopping potential. Vp/Ve.
A parallel-plate capacitor has charge of magnitude 9.00F on each plate and capacitance 3.00F when there is air between the plates. The plates are separated by 2.00 mm. With the charge on the plates kept constant, a dielectric with =5 . is inserted between the plates, completely filling the volume between the plates, (a) What is the potential difference between the plates of the capacitor, before and after the dielectric has been inserted? (b) What is the electrical field at the point midway between the plates before and after the dielectric is inserted?
Can a contour map help you visualize the electric potential energy? Explain.