Please answer part B. Thanks :) Problem 19.101 The figure shows an electron ente

Question

Please answer part B. Thanks :)

Problem 19.101 The figure shows an electron entering a parallel-plate capacitor with a speed of 5.55x106 m/s.The electric field of the capacitor has deflected the electron downward by a distance of 0.618 cm at the point where the electron exits the capacitor. (Figure 1) Figure 1 of 1 2.25 cm 0.618 cm Part A Find the magnitude of the electric field in the capacitor E 4280 N/C Submit My Answers Give Up Correct Part B Find the speed of the electron when it exits the capacitor Submit My Answers Give Up m/s

Explanation / Answer

u =Vx=5.55*10^6m/s

Force is only in y direction hence a is also only in y direction

ay=F/m=Q*E/m and deflection in y direction y=1/2*ay*t^2 and L=Vx*t=>t=L/Vx =0.0225/(5.55*10^6)=4.054*10^-9s

=>y =1/2*Q*E/m*L^2/Vx^2 =Q*E*L^2/(2*m*Vx^2)

=>E =y*2*m*Vx^2/(Q*L^2) =0.618*10^-2*2*9.1094*10^-31*(5.55*10^6)^2/(1.6021*10^-19*0.0225^2)=4276.02N/C

b)Vy=Uy+ay*t=0+1.6021*10^-19*4276.02*4.054*10^-9/(9.1094*10^-31)=3.048*10^6m/s

V^2=Vx^2+Vy^2=(5.55*10^6)^2+(3.048*10^6)^2 =4.01*10^13 => V=6.332*10^6m/s

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