Question

The figure below shows a proton entering a parallel-plate capacitor with a speed of 2.20×105 m/s. The proton travels a horizontal distance x = 7.20 cm through the essentially uniform electric field. The electric field of the capacitor has deflected the proton downward by a distance of d = 0.760 cm at the point where the proton exits the capacitor. (You can neglect the effects of gravity.) capacitor.png Using kinematics, find the vertical acceleration (including sign) of the proton in this electric field. Incorrect. Tries 4/8 Previous Tries Find the magnitude of the force on the proton. Tries 0/8 Find the strength of the electric field within the capacitor. Tries 0/8 Find the speed of the proton when it exits the capacitor.

The figure below shows a proton entering a parallel-plate capacitor with a speed of 2.20x105 m/s. The proton travels a horizontal distance x 7.20 cm through the essentially uniform electric field. The electric field of the capacitor has deflected the proton downward by a distance of d = 0.760 cm at the point where the proton exits the capacitor. (You can neglect the effects of gravity.) Using kinematics, find the vertical acceleration (including sign) of the proton in this electric field 6.73e-3mis 2 Submit Answer Incorrect. Tries 4/8 Previous Tries Find the magnitude of the force on the proton Submit Answer Tries 0/8 Find the strength of the electric field within the capacitor. Submit Answer Tries 0/8 Find the speed of the proton when it exits the capacitor. Submit Answer Tries 0/8

Answer 1

"V_0 x = v_0 = 2 . 20 Times 10^5 m/s, delta x = 7 . 20 cm = 0 .0 72m delta y = d = 0 . 760 cm 0 . 0076m delta x = v_0 t t = delta x/v_0 = 0 . 0 72/2.2 Times 10^5 = 3.273 Times 10^-7 5 delta t = v_0 t + 1/2 ay t^2 x (0.0070) + 1/2 ay (3.273 Times 10^-7)^2 (i) x ay Times = 1.42 Times 10 "" m/s^(ii) F = m ay = (1.67 Times 10^- 27) (1 .4 2 Times 10 "") = 2.37 Times 10^- 16 N (iii) F = eE, E = f/e = 2.37"

EDIT:

(i) will be negative

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