A proton and an alpha particle are released from rest at different locations from the negative plate of a charged parallel plate capacitor. The plates are 15 mm apart, and they are charged to a potential difference of 680 V. The alpha particle is placed at the point where the potential is 600 V. Where do you need to place the proton so that both particles reach the negative plate with the same speed

The potential of the point where we need to place the proton so that both particles reach the negative plate with the same speed is 300 V

Explanation:

The Kinetic Energy, KE = The charge of the particle, Q × The voltage of the particle, V

Therefore, we have;

KE =  1/2 × m × v² = Q × V

The given parameters are;

The distance between the plates of the capacitor = 15 mm

The potential difference to which the capacitor is charged = 630 V

The potential of the point at which the alpha particle is placed = 600 V

The mass of the alpha particle ≈ 4 × Mass of the proton = 4 amu

The charge of the alpha particle = +2

The mass of a proton ≈ 1 amu

The charge of a proton = +1

Therefore, for the alpha particle, we have;

KE =  1/2 × 4 × v² = 2 × 600

v² = 2×(600/(2) = 600

v² = 600

For the proton, given that both particles reach the negative plate with the same velocity, we have;

v² for the proton and the alpha particle are equal

Substituting the known values gives;

1/2 × 1 × 600 = 1 ×

= 1/2 × 1 × 600/1 = 300

= 300 V

The potential of the point where we need to place the proton so that both particles reach the negative plate with the same speed, v, is 300 V.

increase masses, decrease separation