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 1.1 cm, and the electric field within the capacitor has a magnitude of 2.2 x 106 V/m. What is the kinetic energy of the electron just as it reaches the positive plate?

3.88 * 10^(-15) J

Explanation:

We know that the Potential energy of the electron at the beginning of its motion is equal to the Kinetic energy at the end of its motion, when it reaches the plates.

First, we get the potential and potential energy:

Electric potential = E * r

E = electric field

r = distance between plates

Potential = 2.2 * 10^6 * 0.011

= 2.42 * 10^4 V

The relationship between electric potential and potential energy is:

P. E. = q*V

q = charge of electron = 1.602 * 10^(-19) C

P. E. = 2.42 * 10^4 * 1.602 * 10^(-19)

P. E. = 3.88 * 10^(-15) J

real, smaller and upside-down.   so the answer should be b.