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Transcript
Movement of Charged Particles Electrostatics Lesson 8 Chapter 11 When an electron is placed near a negatively charged plate it will have high potential energy. This electric force gives the electron kinetic energy. A change in energy can only occur when work is done on an object. W=∆E Work done by the field = ∆ kinetic energy. Fe•d = Ekf - Eki Eqd = 0.5 m (vf2 - vi2) Since E = V/d then V= Ed so Vq = 0.5 m (vf2 - vi2) An electron is placed in an electric field of 1.00 x 104 N/C to the left. The plates are 10.0 mm apart. Determine the change in kinetic energy the velocity with which the electron hits the positive plate the change in potential energy An alpha particle is placed in an electric field. The vertical plates are connected to a 100 V source of potential energy. Calculate the work done by the field on the alpha particle the gain of kinetic energy of the alpha particle. ratio of the work done per charge Electron Volt unit of energy Definition: One electron volt is the energy gained by a particle having a charge of one elementary charge and is accelerated through a 1.0 V potential difference. 1.00 eV = 1.60 x 10-19 J (see data sheet) Suppose that one electron is accelerated through a 9.00 V potential difference. How much energy does the charge now have? show answer in J and eV NOTE: IF the question asks for velocity then you MUST have energy in Joules. A proton, traveling at 1.5 x 106 m/s, enters perpendicular to an electric field between parallel plates that are 2.0 mm apart and charged by a 9.0 V battery. If the proton enters at the positive plate’s edge, how far into the field will it go before it contacts the negative plate? Assignment Read p.570-575 Do p.571 #1,2 p. 573 #1,2 p. 574 #1,2 Review: SNAP p. 90 all