g 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

d₂ = 6.18 mm

Explanation:

We can work on this interesting exercise using the concepts of energy conservation

alpha (1) particle, with have a charge q = 2e

Starting point

            Em₀ = U = q V₁

final point

            = ½ m v²

             Em₀ = Em_{f}

            2eV₁ = ½ m₁ v²

           v² =

proton (2) particle, with have a charge q = e

Starting point

          Em₀ = qV₂

Final point

          Em_{f} = ½ m₂ v²

          Em₀ = Em_{f}

          eV₂ = ½ m₂ v²

          v² =

in the exercise, it should be noted that the two particles have the same velocity when reaching the plate, therefore let us solve the velocity in each equation and equal

        \frac{4 \ e \ V_{1}}{m_{1} }= \frac{2 \ e \ V_{2} }{m_{2} }

the alpha particle is composed of two protons and two neutrons, therefore in first approximation

          m₁ = 4 m₂

subtitute

         V₂ = V₁

let's calculate

         V₂ = 600

         V₂ = 300 V

To find the distance we use the relationship between the electric field and the potential difference

         V = -d E

The electric field between the plates is constant, so

          E = - V / d = - V₂ / d₂

          d₂ =

let's calculate

          d₂ =  

          d₂ = 6.18 mm

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray