The figure shows an electron entering a parallel-plate capacitor with a speed of

Question

The figure shows an electron entering a parallel-plate capacitor with a speed of v = 5.65 ? 106 m/s. The electric field of the capacitor has deflected the electron downward by a distance ofd = 0.618 cm at the point where the electron exits the capacitor.

The figure shows an electron entering a parallel-plate capacitor with a speed of v = 5.65 ? 106 m/s. The electric field of the capacitor has deflected the electron downward by a distance of a = 0.618 cm at the point where the electron exits the capacitor. Find the magnitude of the electric field in the capacitor. Find the speed of the electron when it exits the capacitor.

Explanation / Answer

The time the electron is between the plates = L/vx = 0.0225/5.9x10^6 = 3.8136x10^-9s

Now the acceleration of the electron is F/m = E*q/m

So y = 1/2*a*t^2 = 1/2*E*q/m*t^2

So E = 2*y*m/(q*t^2) = 2*0.618x10^-2*9.11x10^-31/(1.60x10^-19*(... = 4840N/C

B) vx doesn't change so vx = 5.9x10^6

vy = a*t = Eq/m*t = 4840*1.60x10^-19/9.11x10^-31*3.8136x10^-... = 3.24x10^6m/s

speed = sqrt((5.9x10^6)^2 + (3.24x10^6)^2) = 6.73x10^6m/s

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