A parallel-plate capacitor consists of two plates, each with an area of 29 cm2cm2 separated by 3.0 . The charge on the capacitor is 7.8 nCnC . A proton is released from rest next to the positive plate. Part A How long does it take for the proton to reach the negative plate

   t = 2.09 10⁻³ s

Explanation:

We must solve this problem in parts, first we look for the acceleration of the electron and then the time to travel the distance

let's start with Newton's second law

        ∑ F = m a

the force is electric

        F = q E

we substitute

        q E = m a

        a =

        a =

        a = 1.37 10³ m / s²

now we can use kinematics

        x = v₀ t + ½ a t²

indicate that rest starts v₀ = 0

        x = 0 + ½ a t²

        t =

        t =

        t = 2.09 10⁻³ s

e

explanation:

there is no change in the electric potential energy as it is in rest on both conditions.