A proton enters a parallel-plate capacitor traveling to the right at a speed of

Question

A proton enters a parallel-plate capacitor traveling to the right at a speed of 1.274 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 2.70 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?

A proton enters a parallel-plate capacitor traveling to the right at a speed of 1.274 times 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 2.70 times 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? m

Explanation / Answer

a = 1.6*10^-19*2.7*10^-4/1.67*10^-27 = 25868 m/s2

time spent = sqrt(2*0.00795/25868) secs = 0.000784 secs

Horizontal distance = 1.274 * 10^-5*0.000784 m

= 10*10^-9 m

= 10^-8 m

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