(a)
First, calculate the change in potential energy of the particle as it moves from the positive plate to the negative plate using the formula:
ΔPE = qΔV
Where
q = charge of the particle = 2e = 2 x 1.6 x 10^-19 C
ΔV = potential difference = 2 x 10^3 V
ΔPE = (2 x 1.6 x 10^-19 C) x (2 x 10^3 V)
ΔPE = 6.4 x 10^-16 J
This change in potential energy will be converted to kinetic energy as the particle accelerates towards the negative plate.
K.E. = ΔPE
0.5mv^2 = 6.4 x 10^-16
v = sqrt((2 x 6.4 x 10^-16) / 6.6 x 10^-27)
v = sqrt(1.94 x 10^11)
v = 1.39 x 10^6 m/s
Therefore, the velocity with which the alpha particle reaches the negative plate if it started from rest at the positive plate is 1.39 x 10^6 m/s.
(b)
If the particle starts from rest at a point halfway between the plates, it will already have some initial kinetic energy when it reaches the negative plate.
Since it starts from rest at a point halfway between the plates, the potential energy at that point is half of the potential energy at the positive plate.
So, the change in potential energy at this point is half of the total change in potential energy:
ΔPE = 0.5 x 6.4 x 10^-16 J = 3.2 x 10^-16 J
This change in potential energy will be converted to kinetic energy as the particle accelerates towards the negative plate.
K.E. = ΔPE
0.5mv^2 = 3.2 x 10^-16
v = sqrt((2 x 3.2 x 10^-16) / 6.6 x 10^-27)
v = sqrt(9.69 x 10^10)
v = 3.1 x 10^5 m/s
Therefore, the velocity with which the alpha particle reaches the negative plate if it started from rest at a point halfway between the plates is 3.1 x 10^5 m/s.
An a particle has a positive charge of 2e and a mass of 6.6
1027 kg. With
what velocity would an a particle reach the negative plate of a parallel-plate
apparatus with a potential difference of 2.0
103 V
(a) if it started from rest at the positive plate
(b) if it started from rest at a point halfway between the plates
1 answer