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A newly discovered light positively charged particle has a mass of m and charge q. Suppose it moves within the vicinity of an extremely heavy (fixed in place) particle with a positive charge Q and mass M. When the light particle is xi distance from the heavy particle, it is moving directly away from the heavy particle with a speed of vi. a) What is the lighter particle's speed when it is xf away from the heavy particle? (Consider the Newtonian Gravitation acting between the two charged particles. Ignore the effects of external forces)... (disregard current answers if wrong)

A newly discovered light positively charged particle has a mass of m and charge q. Suppose it moves within the vicinity of an extremely heavy (fixed in place) particle with a positive charge Q and mass M. When the light particle is xi distance from the heavy particle, it is moving directly away from the heavy particle with a speed of vi. a) What is the lighter particle's speed when it is xf away from the heavy particle? (Consider the Newtonian Gravitation acting between the two charged particles. Ignore the effects of external forces)... (disregard current answers if wrong)

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A newly discovered light positively charged particle has a mass of m and charge q. Suppose it moves within the vicinity of an extremely heavy (fixed in place) particle with a positive charge Q and mass M. When the light particle is xi distance from the heavy particle, it is moving directly away from the heavy particle with a speed of vi. a) What is the lighter particle's speed when it is xf away from the heavy particle? (Consider the Newtonian Gravitation acting between the two charged particles. Ignore the effects of external forces)... (disregard current answers if wrong)

Solution: We may solve this using two approaches. One involves the Newton's Laws and the other involving Work-Energy theorem. To avoid the complexity of vector solution, we will instead employ the Work-Energy theorem, more specifically, the Conservation of Energy Principle. Let us first name the lighter particle as object 1 and the heavy particle as object 2. Through work-energy theorem, we will take into account all of the energy of the two-charged particle system before and after traveling a certain distance as KE1F + KE2F + PENewtonianf + Uelasticf + Uelectricf = KE1¡ + KE2¡ + PENewtoniani + Uelastici Uelectrici + Since the heavy particle remains fixed, before and after the motion of the lighter particle, it does not have any velocity, moreover, there is no spring in involved, so KE16 + 0 + PENewtonianf + Uelectricf = KE11 + 0 + PENewtoniani Uelectrici (Equation 1) For all energies, we know the following KE - } mv² Gm,m2 PENewtonian U elastic %3D Uelectric = (1/ 4piepsilon0 ( qQ /(r 2 ))
Transcribed Image Text:Solution: We may solve this using two approaches. One involves the Newton's Laws and the other involving Work-Energy theorem. To avoid the complexity of vector solution, we will instead employ the Work-Energy theorem, more specifically, the Conservation of Energy Principle. Let us first name the lighter particle as object 1 and the heavy particle as object 2. Through work-energy theorem, we will take into account all of the energy of the two-charged particle system before and after traveling a certain distance as KE1F + KE2F + PENewtonianf + Uelasticf + Uelectricf = KE1¡ + KE2¡ + PENewtoniani + Uelastici Uelectrici + Since the heavy particle remains fixed, before and after the motion of the lighter particle, it does not have any velocity, moreover, there is no spring in involved, so KE16 + 0 + PENewtonianf + Uelectricf = KE11 + 0 + PENewtoniani Uelectrici (Equation 1) For all energies, we know the following KE - } mv² Gm,m2 PENewtonian U elastic %3D Uelectric = (1/ 4piepsilon0 ( qQ /(r 2 ))
where in we have m1 = m, m2 = M, q1 = q and q2 = Q By substituting all these to Equation 1 and then simplifying results to = sqrt( v 1 2 + ( ( a V 2 Q ( 2piepsilon0 m ) - GmM/r |) - (1/x ) ) + Take note that capital letters have different meaning than small letter variables/constants.
Transcribed Image Text:where in we have m1 = m, m2 = M, q1 = q and q2 = Q By substituting all these to Equation 1 and then simplifying results to = sqrt( v 1 2 + ( ( a V 2 Q ( 2piepsilon0 m ) - GmM/r |) - (1/x ) ) + Take note that capital letters have different meaning than small letter variables/constants.
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