Question

A parallel-plate capacitor consists of two plates, each with an area of 29 cm2cm2 separated by 3.0 mmmm. 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

Answer 1

t = 2.09 10⁻³ s

Step-by-step 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 =
`(q)/(m) \ E`

a =
`(1.6 \ 10^(-19))/( 9.1 \ 10^(-31) ) \ 7.8 \ 10^(-9)`

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 =
`\sqrt{(2x)/(a) }`

t =
`\sqrt{\frac {2 \ 3 \ 10^(-3)}{ 1.37 \ 10^3} }`

t = 2.09 10⁻³ s