# An electron is moving horizontally enters a region of constant electric field field produced by a...

## Question:

An electron is moving horizontally enters a region of constant electric field field produced by a parallel plate capacitor with a positive plate on the bottom and the negative plate on the top. The electric field is directed perfectly vertically. The electron enters the field with a speed of {eq}4.55 \times 10^6\ m/s {/eq} and exits the field {eq}0.618\ cm {/eq} lower than it entered.

a) What is the magnitude of the electric field produced by the capacitor?

b) What is the velocity of the electron when it exits the capacitor's field?

## Electron in a Uniform Electric Field

The electron enters the region of a uniform electric field moving perpendicular to the direction of the field. The motion described by the electron inside the capacitor is like projectile motion, we can analyze it by dividing it between the horizontal motion, along the {eq}x {/eq} axis and a perpendicular motion along the {eq}y {/eq} axis. The electric force is along the {eq}y {/eq} axis, hence, in the horizontal direction the electron is moving with constant velocity, and in the vertical direction the electron moves with constant acceleration under the action of the constant electric field.

## Answer and Explanation:

The situation is sketched in the figure below.

The electron initial velocity is only in the positive {eq}x {/eq} direction,

{eq}v_{0x}=4.55\times 10^6 \; \rm m/s {/eq},

{eq}v_{0y}=0 {/eq}.

As the positive plate is at the bottom, the electric force on the electron is, as shown in the figure, in the positive {eq}y {/eq} axis, according to our selection of the coordinate system.

Analyzing the motion separately along the {eq}x {/eq} and {eq}y {/eq} axis, we have for the {eq}x {/eq} axis,

{eq}\Delta x=v_{0x}\Delta t {/eq},

where {eq}\Delta x {/eq} is the distance traveled by the electron before it leaves the electric field.

Assuming {eq}\Delta x=10\; \rm cm {/eq}.

Then we can obtain the time the electron is traveling inside the capacitor,

{eq}\Delta t=\dfrac{\Delta x}{v_{0x}}=\dfrac{10\times 10^{-2}\; \rm m}{4.55 \times 10^6 \; \rm m/s}=2.2\times 10^{-8}\; \rm s {/eq}.

In the {eq}y {/eq} axis the particle travels {eq}h=0.618\; \rm cm {/eq}.

{eq}h=v_{0y}t+\dfrac{a\Delta t^2}{2}=\dfrac{a\Delta t^2}{2} {/eq}.

The acceleration of the electron inside the capacitor is,

{eq}a=\dfrac{2h}{\Delta t^2}=\dfrac{2\times 0.618\times 10^{-2}\; \rm m}{2.2^2\times 10^{-16}\; \rm s^2}=2.6\times 10^{13}\; \rm m/s^2 {/eq}.

The electric force is the only force acting on the electron, using Newton's second law and knowing the acceleration we can find the electric field,

{eq}qE=ma {/eq},

{eq}E=\dfrac{ma}{q}=\dfrac{9.1\times 10^{-31}\; \rm kg \cdot 2.6\times 10^{13}\; \rm m/s^2}{1.6 \times 10^{-19}\; \rm C}=148\; \rm V/m {/eq}.

The electric field between the plates of the capacitor if we assumed that the distance traveled by the electron is {eq}\Delta x =10\; \rm cm {/eq} is {eq}\boxed{E=148\; \rm V/m} {/eq}.

b) To find the final velocity of the electron when it leaves the capacitor we need to find the component of the velocity along the {eq}y {/eq} axis.

Using the kinematic equation for motion with constant acceleration,

{eq}a=\dfrac{v_{fy}^2-v_{0y}^2}{2\Delta x} {/eq},

where {eq}v_{0y}=0 {/eq}, we have,

{eq}v_{fy}=\sqrt{2a\Delta x}=\sqrt{2\cdot 2.6\times 10^{13}\; \rm m/s^2\cdot 0.10\; \rm m}=2.3\times 10^6\; \rm m/s^2 {/eq}.

Hence, the velocity of the electron when it exits the field is,

{eq}v_f=\sqrt{v_{0x}^2+v_{fy}^2}\approx 5.1\times 10^6\; \rm m/s {/eq}.

The final velocity of the electron is {eq}\boxed{ 5.1\times 10^6\; \rm m/s} {/eq}.

#### Learn more about this topic:

Electric Field & the Movement of Charge

from AP Physics 2: Exam Prep

Chapter 8 / Lesson 3
31K