Suppose a proton enters a linear accelerator at a speed of 1.00 times 106 ms. Th

Question

Suppose a proton enters a linear accelerator at a speed of 1.00 times 106 ms. The linear accelerator consists of two charged plates 4.00 cm apart, each with an opening for the proton. The proton exits the linear accelerator at a speed of 4.00 times 106 m/s. A) What is the change of kinetic energy of the proton? What is the change of electric potential energy of the proton? What is the electric potential difference from where the proton enters to where it exits? What is the magnitude of the electric field between the plates, assuming it is uniform?

Explanation / Answer

a)change in kinetic energy=1/2 m v^2 =1/2*1.6*10^-27*10^12 =8*10^-16 joules. b)change in electric potential energy=8*10^-16 joules(equal to change in kinetic energy) c)eV=8*10^-16 or V=5000V d)electric field between the plates= V/d =5000/4*10^-2 =125000 N/C hope my answer i correct and you will appreciate my help by awarding me karma points.

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