Question

Beams of high-speed protons can be produced in "guns" using electric fields to accelerate the protons. Find the acceleration would a proton experience if the gun's electric field were 2.72 × 10⁴ N/C. What speed would the proton attain if the field accelerated the proton through a distance of 1.17 cm? (enter speed for answer).

Answer 1

Answer:

Given that, magnitude of electric field E = 2.72 x 10⁴ N/C,

The proton will experience an acceleration in the external electric field E, that means the electric force from the field E creates the acceleration on the proton, therefore F_e = ma.

The electric force on a charged particle moving in a electric field is F_e = eE

Then eE = ma   or a = eE/m

Therefore, a = (1.6 x 10^-19C) (2.72 x 10⁴ N/C) / (1.67 x 10^-27 kg) = 2.60 x 10¹² m/s².

Using the expression v² - u² = 2as

Here, initial speed of proton is zero so u = 0 m/s

Therefore. final speed v = (2as)^1/2 = [ 2(2.60 x 10¹² m/s²)(1.17 x 10^-2 m) ]^1/2 = 2.46 x 10⁵ m/s.