The figure below shows an electron entering a parallel-plate capacitor with a speed of v = 5.70×106 m/s.

1. The figure below shows an electron entering a parallel-plate capacitor with a speed of v = 5.70×106 m/s. The electric field of the capacitor has deflected the electron downward by a distance of d = 0.590 cm at the point where the electron exits the capacitor. Find the magnitude of the electric field in the capacitor. in N/Chttps://general.physics.rutgers.edu/gifs/CJ/18-13.gif

The time the electron takes between the plates = L / Vx = 0.0225/5.7 x10^6 = 3.94×10^-9sThe acceleration of the electron is F/m = E*q/mSince y = 1/2*a*t^2 = 1/2*E*q/m*t^2So E = 2*y*m/(q*t^2)…