An electron is released at he negative plate of a parallel platecapacitor and ac

Question

An electron is released at he negative plate of a parallel platecapacitor and accelerates to the positive plate. (a) As theelectron gains kinetic energy, does its electric potential energyincrease or decrease? Why? (b) The difference in theelectron's electric potential energy between the positive andnegative plates is EPEpositive -EPEnegative. How is this difference related to thecharge on the electron (-e) and to the differenceVpositive - Vnegative in theelectric potential between plates? (c) How is the potentialdifference Vpositive -Vnegative related to the electric field within thecapacitor and the displacement of the positive plate relative tothe negative plate?
Problem: The plates of a parallel plate capacitor areseparated by a distance of 1.2 cm, and the electric field withinthe capacitor has a magnitude of 2.1 X 106V/m. Anelectron starts from rest at the negative plate and accelerates tothe positive plate. WHat is the kinetic energy of theelectron just as the electron reaches the positive plate?
Any explanation would be greatly appreciated!
Problem: The plates of a parallel plate capacitor areseparated by a distance of 1.2 cm, and the electric field withinthe capacitor has a magnitude of 2.1 X 106V/m. Anelectron starts from rest at the negative plate and accelerates tothe positive plate. WHat is the kinetic energy of theelectron just as the electron reaches the positive plate?
Any explanation would be greatly appreciated!

Explanation / Answer

(a)    the electric potential energydecreases    the electric force F is aconservative force, so the total energy (kinetic energy pluselectric potential energy) remains    constant as the electron moves across thecapacitor    so as the electron accelerates and itskinetic energy increases, its electric potential energydecreases (b)    from the theory we have the equation    VB - VA= (EPEB / qo) - (EPEA /qo)                  = - WAB / qo    so the change in the electron’selectric potential energy is equal to the charge on the electron(-e) times the potential    difference between the plates    EPEpositive -EPEnegative = (-e)(Vpositive -Vnegative) (c)    the electric field E is related to thepotential difference between the plates and thedisplacement s by    E = - V / s       = - (Vpositive- Vnegative) / s    note that (Vpositive -Vnegative) and s are positive numbers, so theelectric field is a negative number, denoting that it    points to the left    so the change in the electron’selectric potential energy is equal to the charge on the electron(-e) times the potential    difference between the plates    EPEpositive -EPEnegative = (-e)(Vpositive -Vnegative) (c)    the electric field E is related to thepotential difference between the plates and thedisplacement s by    E = - V / s       = - (Vpositive- Vnegative) / s    note that (Vpositive -Vnegative) and s are positive numbers, so theelectric field is a negative number, denoting that it    points to the left

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