The protons initially are located where the electric potential has a value of 7.60 mv and then they travel through a vacuum to a region where the potential is zerdo. (a) find the final speed of these protons m/s (b) find the accelerating electric field strength if the potential changed uniformly over a distance of 1.70 m. mv/m

(a) 3.82 x 10⁷ m/s

(b) 4.5 MV/m

Explanation:

(a)

ΔV = change in the electric potential as the proton moves = 7.60 x 10⁶ Volts

q = magnitude of charge on proton = 1.6 x 10⁻¹⁹ C

v = speed gained by the proton

m = mass of proton = 1.67 x 10⁻²⁷ kg

Using conservation of energy

Kinetic energy gained by proton = Electric potential energy

(0.5) m v² = q ΔV

inserting the values

(0.5) (1.67 x 10⁻²⁷) v² = (1.6 x 10⁻¹⁹) (7.60 x 10⁶)

v = 3.82 x 10⁷ m/s

(b)

d = distance over which the potential change = 1.70 m

Electric field is given as

E = ΔV/d

E = 7.60 x 10⁶/1.70

E = 4.5 x 10⁶ V/m

E = 4.5 MV/m

15.3 cos 65 degrees trig