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E&M Capacitor Problem Stephanie posted on 2017-01-31 14:56:34 |
Two parallel plates 1.10 cm apart are equally and oppositely charged. An electron is released from rest at the surface of the negative plate and simultaneously a proton is released from rest at the surface of the positive plate. How far from the negative plate is the point at which the electron and proton pass each other? |
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Replied by Iris on 2017-02-02 21:34:53 |
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This question is kind of tricky because you are mainly working with forces and kinematics. In this situation, the force acting on the electron and proton from the electric fields are the same because electric field between the two capacitor plates is uniform. Based on Newton's second law, F=ma. Since F of electron and F of proton is the same, you can equate them and get -Mp*Ap= Me*Ae (negative acceleration of proton if we assume its going in the opposite direction of the coordinate system) isolate: Ap = -Me/Mp * Ae From the question, we know that the electron Yei=0, Vie=0 while the proton Ypi = d, Vip=0. Final position and t are equal since that is what were looking for. Set up the kinematic equation: Yf = Yi +Viy*t + 1/2*a*t^2 for both electron and proton, equate them,plug the Ap equation, isolate t, and plug t back into previous kinematics equation and voila! Basically just do system of equations. Your answer should be 1.10cm (which makes sense since the proton is much bigger than the electron, it will barely move.) Hope that helps! |