Question

A proton enters a parallel-plate capacitor traveling to the right at a speed of 1.276 x 10^-5 m/s, as shown in the figure. The distance between the two plates is 1.59 cm. The proton enters the capacitor halfway between the top plate and the bottom plate; that is, a distance r = 0.795 cm from each plate, as shown in the figure. The capacitor has a 2.75 x 10^-4 N/C uniform electric field between the plates that points downward from the top plate to the bottom plate. Neglecting gravitational forces, what horizontal distance does the proton traverse before the proton hits the bottom plate?

Answer 1

acceleration acting on proton

a = E q / m

a = 2.75 10^-4* 1.6*10^-19 / (1.67*10^-27)

a = 26347.3 m/s^2

Considering motion along vertical

y = 0.5 at^2

0.795*10^-2 = 0.5* 26347.3* t^2

t = 7.768*10^-4 second

Horizontal distance travelled till this time

x = vt

x = 1.276*10^-5*7.768*10^-4

x = 9.91*10^-9 m

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