Find velocity of the electron when it comes out of the parallel plates
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2- An Electron is emitted by heating a cathode and is accelerated by the potential difference of V volts between the anode and a cathode. The electron emerges out from the anode with velocity 45109311.4 m/s. Find velocity of the electron when it comes out of the parallel plates. Use x = 17 m and y = 0.05 m.
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- 1. Adjust the voltage to 0.05x V and adjust the plate area to 10y mm² and the separation distance to 0.2z mm. What is the magnitude of the charge (in Coulombs) stored in each plates for x= −27, y= 21, and z= 39? 2. Adjust the voltage to 0.05x V and adjust the plate area to 10y mm² and the separation distance to 0.2z mm. What is the magnitude of the Potential Energy in Joules in capacitor for x= -8 and y= 10? 3. Adjust the voltage to 0.05x V and adjust the plate area to 10y mm² and the separation distance to 0.2z mm. What is the magnitude of the charge stored in each plates for x= −25 and y= 14?c) Considering electron and proton as two charged particles separated by d = 5 × 10-11 m calculate the following: i) the electrostatic potential at a distance d from the proton. ii) the potential energy of the electron in the field of the proton iii) the Coulomb force between proton and electron iv) the gravitational force between the proton and electron and find its ratio to the Coulomb force. Take the mass of the proton 1.7 x 10-27 kg, the mass of the electron 9.1 x 10-31 kg, the electron charge -1.6 x 10-19 C, the value of the universal gravitational constant 1 6.7 × 10-11 N kg-2m-2 and Απο 9 x 10° m/F.2) Consider a beam of electron so that each has a kinetic energy of 1.95 x 10-15 J. We want to set up an electric field parallel to the beam that stops the electrons after they've crossed a distance of 0.14m? What is the electric field strength that we need?
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