Answer:
(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
