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- 2. Charged particle interactions and energy conservation A stationary charge, Q = + 31.0uC, remains fixed in position. A moving charge, q = + 7.70µC, with mass, m = 1.18 x 10 kg, heads toward this stationary charge along the x axis. The moving charge starts at position A with an initial speed of 892 m/s. Point A is 1.64 m from the fixed charge. Point B is 0.550 m from the fixed charge. a. Use energy conservation principles to set up an energy conservation equation for this system. b. Solve the energy conservation equation for the speed of the moving charge, vx, when it reaches point B. q. m >+ x axis Fixed stationary chargeA proton's speed as it passes point A is 4.00×104 m/s. It follows the trajectory shown in the figure. What is the proton's speed at point B?The distance, r, between the proton and electron in the hydrogen atom is about 10^-10m. i) Calculate the electrostatic force between the electron and proton. ii) Determine the value of the potential energy U of the electron in the hydrogen atom. Express your result in electron-volts (eV). Recall 1 eV = 1.6 × 10^−19 J. iii) Determine the value of the kinetic energy of the electron. Express your result in eV. Please answer all parts.
- 1. An electron is to be accelerated in a uniform electric field having a strength of 2.1-106 (a) What energy in keV is given to the electron if it is accelerated through 0.3 m? AKE = keV m (b) Over what distance (in km) would it have to be accelerated to increase its energy by 43 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?Before liquid crystal display LCD computer and television screens, the screens were cathode-ray tube CRTS, in which a bean of electrons was shot towards a Phosphorescent screen at high speed and deflected by electric fields in order to direct the beam to the right place on the screen so that it would form the picture. In a typical television, the speed of the electrons is 1 x10^8 m/s. If the television set is placed so that the CRT beam is horizontal heading east in a place where the earths magnetic field is 5×10^-5 and points horizontally northward what will be the magnetic force on the electron?= A point charge q₁ 9.1 µC is held fixed at 0.42 μC and a mass origin. A second point charge 2 3.2 x 104 kg is placed on the x-axis, 0.96 m from the origin. The second point charge is released at rest. What is its speed when it is 0.24 m from the origin? =