consider the motion of the proton between the plates
d = distance traveled by the proton = distance between the two plates = 1.70 cm = 0.0170 m
v₀ = initial velocity of the proton = 0 m/s
t = time taken to reach the opposite plate = 1.48 x 10⁻⁶ sec
a = acceleration of the proton
using the kinematics equation
d = v₀ t + (0.5) a t²
inserting the values
0.017 = 0 ( 1.48 x 10⁻⁶) + (0.5) a ( 1.48 x 10⁻⁶)²
a = 1.6 x 10¹⁰ m/s²
m = mass of proton = 1.67 x 10⁻²⁷ kg
q = charge on the proton = 1.6 x 10⁻¹⁹ C
E = electric field between the plates
force on the proton due to electric field is given as
F = ma
electric force is given as
F = qE
combining the two equations we get
qE = ma
(1.6 x 10⁻¹⁹) E = (1.67 x 10⁻²⁷) (1.6 x 10¹⁰)
E = 167 N/C
v = final speed of electron as it strikes the negative plate
Using the kinematics equation
v = v₀ + at
v = 0 + (1.6 x 10¹⁰) (1.48 x 10⁻⁶)
v = 23680 m/s
