An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see the drawing). The plates are separated by a distance of 1.8 cm, and the electric field within the capacitor has a magnitude of 2.6 x 106 V/m. What is the kinetic energy of the electron just as it reaches the positive plate?

      K = 4.2 10⁻¹³ J

Explanation:

For this exercise we will use the principle of conservation of energy

starting point. Where the electron is released

          Em₀ = U = e V

final point. On the positive plate

          Em_f = K = ½ m v²

as there is no friction, energy is conserved

          Em₀ = Em_f

          e V = ½ m v²

         K = e V

  let's calculate

         K = 1.6 10⁻¹⁹   2.6 10⁶

         K = 4.2 10⁻¹³ J

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s