Part A Cathode-ray tubes (CRTS) are often found in oscilloscopes and computer monitors. In (Figure 1) an electron with an initial speed of 6.70x106 m/s is projected along the axis midway between the deflection plates of a cathode-ray tube. The potential difference between the two plates is 24.0 V and the lower plate is the one at higher potential. What is the magnitude of the force on the electron when it is between the plates? Express your answer in newtons. x" x•10" Figure 1 of 1 F = N Submit Previous Answers Request Answer X Incorrect; Try Again; 8 attempts remaining 2.0 cm Part B K6.0 cm 12.0 cm -K
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- Problem 18.17 - Enhanced - with Solution You may want to review (Page). For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Parallel plates and conservation of energy. Part A An electron is to be accelerated from a velocity of 3.50x 10 m/s to a velocity of 7.50x105 m/s. Through what potential difference must the electron pass to accomplish this? for Panze for Partido for Part redo foart A refor Part A keyboard shortcuts for Part A help for Part. A Viniital - Viinal= Submit Part B Request Answer Through what potential difference must the electron pass if it is to be slowed from 7.50x105 m/s to a halt? Vinitial - Vinal= Submit for Partfondo for Part redo folet B reor Part B keyboard shortcuts for Part B help for Part B Request Answer VI am stuck on this physics homework, any help would be great!A flower--a large conducting mass on top of a narrow stem--and the ground together act like a capacitor. The flower is one electrode, the earth is the other. A typical value of the capacitance is 0.80 pF . The electric field of the earth induces a charge on the flower and an opposite charge on the ground below. Part A If a flower carries a charge of magnitude 40 pC , what is the approximate potential difference between the flower and the ground below?
- - A proton with initial speed vi =2 x104m/s moves from a 5000V equipotential to a 4090V equipotential as shown. 5000v 4990V Vq =? Part A What is the initial Kinetic Energy of the proton? ? KE = Joules Submit Request AnswerThe figure shows an electron entering a parallel-plate capacitor with a speed of 5.7×106 m/s. The electric field of the capacitor has deflected the electron downward by a distance of 0.618 cm at the point where the electron exits the capacitor. (Figure 1) Figure V + + + -2.25 cm- + + É + + + < 1 of 1 0.618 cm + Part A Find the magnitude of the electric field in the capacitor. 15| ΑΣΦ 3 E- 3662.75 ! Your answer should not contain commas. No credit lost. Try again. Part B Submit Previous Answers Request Answer www ||| ΑΣΦ v=6.36• 106 Submit Find the speed of the electron when it exits the capacitor. Provide Feedback www. ? Previous Answers Request Answer ? N/C m/s X Incorrect; Try Again; 29 attempts remaining Review your calculations and make sure you round to 3 significant figures in the last step.The electric field strength is 4.50x104 N/C inside a parallel-plate capacitor with a 1.80 mm spacing. A proton is released from rest at the positive plate. Part A You may want to review (Pages 688 - 690). What is the proton's speed when it reaches the negative plate? Express your answer with the appropriate units. Value Units
- d and eTwo spherical shells have a common center. The inner shell has radus R₁5.00 cm and charge g+5.00×10 C: the outer shell has radius R-15.0 cm and charge g2-7.00x10- C. Both charges are spread uniformly over the shell surface. Take V-0 at a large distance from the shells Part A What is the electric potential due to the two shells at the distancer-2.50 cms from their common center Express your answer with the appropriate units. HAO? Value UnitsI know that part a is 0 V. I tried 3.54x10^6 V for part b but that is incorrect.