A proton starts out moving horizontally from a positive plate to a negative plat

Question

A proton starts out moving horizontally from a positive plate to a negative plate of equal size and equal magnitude of charge. The distance between the two plates is 2 m and the voltage between the two plates is 167.26 V. After a very short time the proton collides with the negative plate at a height lower than the one it was at near the positive plate. Calculate this difference of heights in meters.

Explanation / Answer

the acceleration a = F/m = qE/m = qVd/m NOw using kinematics eqn S=at^2/2 (initial vel = 0) we get d = qVdt^2/2m => t = sqrt(2m/qV) = sqrt( 2*1.672×10-27/(1.6x10^-19x167.26)) = 11.178 x 10^-6 seconds Hope you liked my detailed explanation :)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.