Answer:
= 2.52 x 10^ 6 m/s
Explanation:
The force that acts on charged particles between capacitor plates =
F = (q) (Δv) ÷ d
Here, d = distance between the two plates
q = charge of the charged particle
Δv = voltage
Normally, the force that makes both proton and electron released from rest, giving the charge acceleration is F=m X a. where m= mass and a = acceleration
Poting this equation with the first one, we have:
m X a = (q) (Δv) ÷ d
So, the acceleration of a proton when moving towards a negatively charged plate is
a = (q) (Δv) ÷ (d) (m) {proton}
Likewise, the acceleration of an electron when moving towards a positively charged plate is
a = (q) (Δv) ÷ (d) (m) {electron}
Dividing the proton acceleration formula by the electron acceleration formula we have:
a (proton) / a (electron) = m (proton) / m(electron)
inserting equation of motion to get distance, s
s = ut + 1/2 at^2
recall that electron travel distance, d/2
d/2 = 1/2 at^2
making t the subject of the formula
we have, t =√(d ÷ a(electron))
The distance of proton:
d/2 = ut + 1/2 at^2 [proton}
put d/2 = ut + 1/2 at^2 [proton} into t =√(d ÷ a(electron))
Initial speed, ui = √(d ÷ a(electron)) = (d/2) - (1/2) x (d) (a(proton) + a(electron))
since acceleration wasn't given in the question, lets use mass(elect
ron) ÷ mass(proton) rather than use (a(proton) + a(electron))
Therefore, intial speed= 1/2√((e X Δv) ÷ m(electron)) (1- m(electron)/ m(proton))
Note, e = 1.60 x 10^-19
m(electron) = 9.11 X 10^-31
m(proton) = 1.67 X 10^-27
Input these values into the formula above, initial speed, UI =
= 2.52 x 10^ 6 m/s
