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velocity of electron in parallel plate capacitor

The distance between the plates is 16.8 cm, and the voltage difference is 130 kV. See the answer. To enable Verizon Media and our partners to process your personal data select 'I agree', or select 'Manage settings' for more information and to manage your choices. At speeds higher than > 1/10c (10% speed of light), we have to use relativity for accuracy. This gives K f = 1/2 mv 2 = q ΔV so: v = [ 2qΔV /m ] 1/2. An electron is emitted from the surface of the 0 volt plate, with negligible velocity. • An accelerating voltage varying as V = at, where a is a constant is applied between the plates. 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We and our partners will store and/or access information on your device through the use of cookies and similar technologies, to display personalised ads and content, for ad and content measurement, audience insights and product development. An electron is accelerated inside a parallel plate capacitor. A particle of mass 1 × 1 0 − 2 6 k g and charge 1. Performance & security by Cloudflare, Please complete the security check to access. ( 9m2eal. This problem has been solved! … (The rest mass of the electron is 9.11×10-31 kg, the rest energy of the electron is 511 keV.). 10 -31 kg. Join Yahoo Answers and get 100 points today. U i + K i + W nc = U f + K f. There are no non-conservative forces acting, the initial kinetic energy is zero, and we can define the plate where the charge exits the capacitor as the zero of potential energy (i.e., U f = 0). Yahoo is part of Verizon Media. What is the final velocity β of the electron if you use relativistic mechanics? The separation between the plates is l. The velocity of the electron at the moment it reaches the opposite plate will be : A . KE = (Y-1)mc^2 use this eqn to find gamma (Y) and use v = sqrt[1-(1/Y^2)]c to find the velocity. You may need to download version 2.0 now from the Chrome Web Store. Find out more about how we use your information in our Privacy Policy and Cookie Policy. There are two parallel plates acting as a capacitor, arranged in the y-z plane, one at 0 volts and one at +V volts. Determine the final velocity β of the electron using classical mechanics. An electron is accelerated inside a parallel plate capacitor. | Yahoo Answers. 4 × 1 0 3 k ^ N C − 1 and B = 8 × 1 0 − 2 j ^ W b m − 2, the direction of motion of the particles is physics. Information about your device and internet connection, including your IP address, Browsing and search activity while using Verizon Media websites and apps. if E = − 1 0 2. The figure shows an electron entering a parallel-plate capacitor with a speed of 5.45×106m/s . An electron is launched at a 45? A uniform magnetic field of flux density B field acts parallel to these plates in the z direction. 2 8 × 1 0 6 m / s along the positive X-axis enters a region in which a uniform electric field E and a uniform magnetic field of induction B are present. . Still have questions? The electron leaves the negative plate with a negligible initial velocity and then after the acceleration it hits the positive plate with a final velocity β. equation of motion in y-direction. Determine the final velocity β of the electron using classical mechanics. Convert eV to Joules using the fact that 1 ev = 1x10^-19 J, Y = [(130000ev)(1.60x10^-19 J/ev)]/[(9.11x10^-19)(3x10^8)^2] + 1 = 1.254, v = sqrt{1-[1/(1.254)]}c = 1.81x10^8 m/s = .603c = β. Get your answers by asking now. The electron leaves the negative plate with a negligible initial velocity and then after the acceleration it hits the positive plate with a final velocity β. v= velocity = β, V=voltage diff, KE = kinetic energy, Y = gamma, m = mass, and q = charge, v = sqrt(-2qV/m) use this equation and solve for v. It is derived from the fact that KE = Vq, so, v = sqrt[-2(-1.6x10^-19)(130000)/ (9.11x10^-31 )] = 2.14x10^8 m/s = .712c = β. Your IP: 37.187.171.105 Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. The distance between the plates is 16.8 cm, and the voltage difference is 130 kV. • At the moment t = 0, an electron leaves one plate of a parallel-plate condenser with a negligible velocity. Electron entering a parallel-plate capacitor? 6 × 1 0 − 1 9 C travelling with a velocity 1. angle and a speed of 5.0×106m/s from the positive plate of the parallel-plate capacitor shown in the figure (Figure 1) . You can change your choices at any time by visiting Your Privacy Controls. y(t) = 1 2 ay ⋅t2 (1) Another way to prevent getting this page in the future is to use Privacy Pass. The top of your head has a surface area of 25 cm x 25 cm.

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