A proton is fired with a speed of vi=178260.0m/s from the midpoint of the capacitor shown towards the positive plate. If V2=590.0V, What is the proton's speed as it collides with the negative plate?
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A proton is fired with a speed of vi=178260.0m/s from the midpoint of the capacitor shown towards the positive plate. If V2=590.0V, What is the proton's speed as it collides with the negative plate?
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- A negatively charged particle of mass 8.0 x 10-13 kg is traveling rightward between two plates separated by a distance d = 50 cm, as shown below. The particle is launched with a speed of 1.0 x 104 m/s at the left plate, and strikes the right plate at a speed of 5.0 x 10³ m/s. The left plate is at electric potential V₁ = +1.0 V, while the right plate is at electric potential V₂ = -5.0 V. Assume the electric field between the plates is uniform, and ignore the effect of gravity. 2 (a) (b) V₁ d 7 V₂ Does the potential energy of the particle increase or decrease as it moves from the left plate to the right plate? Is the electric field between the plates directed rightward or leftward? Justify your answers briefly. Show clearly that the electric field between the plates has magnitude 12 V/m (or 12 N/C). Then find the charge of the particle (with correct magnitude and sign). (Note: You may obtain the charge using either energy considerations or a "force and kinematics" approach.The difference in electric potential between the accelerating plates in the electron gun of a TV tube is 25 kV, while the distance between them is 1.5 cm. The magnitude of the uniform electric field between the plates is 3.8 10 2 v/m. 5.6 10² v/m. O 1.7 106v/m. 11'108 v/m.As shown in Figure 1, two parallel plates have a potential ditterence of 12 V and are held a distance of 0.001m apart. A free charge held near the positive plate moves to the right, as shown, when it is released. It then moves through an opening in the negative plate on the right. 12V (a) Is the particle a proton or an electron? (b) What is the energy (in Joule) as it leaves the right plate? (c) How fast is it moving when it reaches the negative plate? (d) What are the magnitude and the direction of the electric field between the plates?
- An electron is to be accelerated in a uniform electric field having a strength of 2.106 (a) What energy in keV is given to the electron if it is accelerated through 0.45 m? AKE = keV m (b) Over what distance (in km) would it have to be accelerated to increase its energy by 45 GeV? d = ✔km Hint: How is potential energy, PE, gained by an electron related to the uniform electric field? How is the potential difference, V, related to the uniform electric field?Answers: a) 3.3 C c/m² 6. Two electrons are fixed 2.00 cm apart. Another electron is shot from infinity and comes to rest midway between the two. What was its initial speed? Answer: 318 m/sAn electron initially at rest, is allowed to accelerate through a potential difference of 1 V, gaining kinetic energy KEe, whereas a proton, also initially at rest, is let accelerate through a potential difference of 1 V, gaining kinetic energy KEp. As ∣qe∣ = ∣qp∣ but mp >> me, therefore, Group of answer choices KEe >> KEp KEe << KEp KEe = KEp All we can say, is KEe ≠ KEp
- An electron is fired at a speed yo = 4.5 x 106 m/s and at an angle D + + + + + + + + Path of the electron 0 00 d X + AV = -45° halfway between two parallel conducting plates that are D = 4.0 mm apart, as in the figure below. The voltage difference between the plates is AV = 135 V. (a) Determine how close, d, the electron will get to the bottom plate. mm (b) Determine the horizontal position where the electron will strike the top plate. (Report the distance from the origin along the x-axis.) mm10. An electron is accelerated from rest through a potential difference of 2.5 x 10° V. (a) What is the relativistic kinetic energy of the electron? (b) What is the speed of the electron? (c) What is the total relativistic energy of the electron? Given that the magnitude of the charge of the electron is 1.602 × 10-19 C and the mass of the electron is 9.11 × 10-3! kg.If an electron is accelerated from rest through a potential difference of 1500 V, what speed does it reach? (e = 1.60 × 10-19 C, melectron = 9.11 × 10-31 kg) 2.3 × 107 m/s 1.5 × 107 m/s 5.3 × 1014 m/s 1.9 × 107 m/s 2.3 × 1014 m/s
- Question 7 b(ii) , b(iii) and 7 a(ii) , 7a(i)The uniform electric field between two charged parallel plates has a magnitude of 4400 V/m (note that 1 V/m = 1 N/C). If the plates are separated by 0.0100 m and an electron is released from the far negative side of the gap, what will the electron's kinetic energy (in eV) and speed be when it just reaches the positive plate? Note that the electron charge = -1.602 x 10-19 C and the electron mass = 9.11 x 10-31 kg. KE V= ev m/sHow much work (in eV) does it take to move a 3 protons through a 30 V potential difference?