Question

Beams of high-speed protons can be produced in "guns" using electric fields to accelerate the protons. (a) What acceleration would a proton experience if the gun's electric field were 2.95 × 104 N/C? (b) What speed would the proton attain if the field accelerated the proton through a distance of 1.26 cm?

Answer 1

(A) the acceleration experienced by the proton 2.821 x 10¹² m/s²

(B) the speed of the proton is 2.67 x 10⁵ m/s

Step-by-step explanation:

Given;

electric field experienced by the proton, E = 2.95 x 10⁴ N/C

charge of proton, Q = 1.6 x 10⁻¹⁹ C

mass of proton, m = 1.673 x 10⁻²⁷ kg

distance moved by the proton, d = 1.26 cm = 0.0126 m

(a)

The force experienced by the proton is calculated as;

F = ma = EQ

where;

a is the acceleration experienced by the proton

`a = (EQ)/(m) \\\\a = (2.95* 10^4 \ * \ 1.6* 10^(-19))/(1.673 * 10^(-27)) \\\\a = 2.821 * 10^(12) \ m/s^2`

(b) the speed of the proton is calculated;

v² = u² + 2ad

v² = 0 + (2 x 2.821 x 10¹² x 0.0126)

v² = 7.109 x 10¹⁰

v = √7.109 x 10¹⁰

v = 2.67 x 10⁵ m/s