An electron (charge -e) moving at velocity v passes through two vertically stack
ID: 1461267 • Letter: A
Question
An electron (charge -e) moving at velocity v passes through two vertically stacked square parallel plates with dimension a on one side, and distance d between plates. A potential difference V is applied across the plates such that the top plate is positively charged. A uniform magnetic field B is applied parrallel to the plates and transverse to the velocity of the electron to deflect the electron such that it does not go into the plates. Part A -
Assuming that the magnetic field is applied only through the volume of the capacitor, what must its magnitude be if the electron strikes a screen, situated a distance L from the nearest side of the capacitor, a distance y above dead center. Dead center of the screen is where the electron would hit if its trajectory were unaltered by either the capacitor or magnetic field.
Hint: Use the kinematic equations to determine the relationship between the net force and the trajectory of the electron: yf=yi+vyit+12ayt2 and xf=xi+vxit+12axt2
Dead center of the screen is where the electron would hit if its trajectory were unaltered by either the capacitor or magnetic field.
Hint: Use the kinematic equations to determine the relationship between the net force and the trajectory of the electron
Dead center of the screen is where the electron would hit if its trajectory were unaltered by either the capacitor or magnetic field.
Hint: Use the kinematic equations to determine the relationship between the net force and the trajectory of the electron: and
B=V/dv +2 ymev/e(a+L)^2 B=V/dv +ymev/e(a+L)^2 B=V/dv +2 ymev/ae(a+2L) B=V/dv +ymev/ae(a+L) B=V/dv 2 ymev/e(a+L)^2 B=V/dv ymev/e(a+L)^2 B=V/dv 2 ymev/ae(a+2L) B=V/dv ymev/ae(a+L)Explanation / Answer
An electron (charge -e) moving at velocity v passes through two vertically stack
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