Controlling the electron beam. An electron is shot through a varying electrical field. Initially, the electron is moving in the \(x\)-direction with a velocity \(v_{x}=100 \mathrm{~m} / \mathrm{s}\). The electron enters the field when it passes the origin. The field varies with time, causing an acceleration of the electron that varies in time: $$ \mathbf{a}(t)=\left(-20 \mathrm{~m} / \mathrm{s}^{2}-10 \mathrm{~m} / \mathrm{s}^{3} t\right) \mathbf{j} . $$ (a) Find the velocity as a function of time for the electron. (b) Find the position as a function of time for the electron. The field is only acting inside a box of length \(L=2 \mathrm{~m}\). (c) How long time is the electron inside the field? (d) What is the displacement in the \(y\)-direction when the electron leaves the box. (We call this the deflection of the electron). (e) Find the angle the velocity vector forms with the horizontal as the electron leaves the box.

Step by step solution

01

Write down the problem information

Initial velocity: \( v_{x}=100 \, \mathrm{m/s} \). Temporal acceleration: \( \mathbf{a}(t)=(-20 \, \mathrm{m/s^{2}} - 10 \, \mathrm{m/s^{3}} \, t) \mathbf{j} \). Length of the field: \( L = 2 \, \mathrm{m} \).

02

Determine the velocity as a function of time (Part a)

Since acceleration is the derivative of velocity with respect to time, integrate the acceleration function to find the velocity. \[ \frac{d\textbf{v}(t)}{dt} = \textbf{a}(t) \] Integrate the acceleration equation: \[ \begin{align*} \textbf{v}(t) &= \textbf{v}_{0} + \textbf{a}(t) \, dt = 100 \, \mathbf{i} + \bigg[ \bigg( -20 \, t \, \mathrm{m/s^{2}} - 5 \, t^{2} \, \mathrm{m/s^{3}} \bigg) \mathbf{j} \bigg] + \textbf{C} \, \]. The constant of integration for the y-component of the velocity will be zero since the initial velocity in the y-direction is zero, therefore: \[ \textbf{v}(t) = 100 \, \mathrm{m/s} \mathbf{i} + (-20 \, t \, \mathrm{m/s^{2}} - 5 \, t^{2} \, \mathrm{m/s^{3}} ) \mathbf{j} \].

03

Determine the position as a function of time (Part b)

Since velocity is the derivative of position with respect to time, integrate the velocity function to get position. \[ \frac{d\textbf{r}(t)}{dt} = \textbf{v}(t) \] Integrating the velocity equation: \[ \textbf{r}(t) = \bigg( 100 \, t \, \mathrm{m/s} \mathbf{i} + \bigg( -10 \, t^{2} \, \mathrm{m/s^{2}} - \frac{5}{3} \, t^{3} \, \mathrm{m/s^{3}} \bigg) \mathbf{j} \bigg) + \textbf{C} \. \] The constant of integration for the y-component will be zero since the initial position in the y-direction is zero from the problem statement, hence: \[ \textbf{r}(t) = 100 \, t \, \mathrm{m/s} \mathbf{i} + \bigg( -10 \, t^{2} \, \mathrm{m/s^{2}} - \frac{5}{3} \, t^{3} \, \mathrm{m/s^{3}} \bigg) \mathbf{j} \, \].

04

Determine the time the electron is inside the field (Part c)

The length of the field is given as \(L = 2\, \mathrm{m}\), and the initial velocity in the x-direction is \(v_{x} = 100 \, \mathrm{m/s}\). The time \(t\) can be found by setting the x-component of position equal to 2 m and solving for \(t\): \[ 100t = 2 \, \mathrm{m} \. \] Therefore, \[ t = 0.02 \, \mathrm{s} \].

05

Find the displacement in the y-direction (deflection) (Part d)

Use the time found in Step 4 to calculate the y-component of the position function. From the position function, \( \textbf{r}(t) = 100 \, t \, \mathbf{i} + \bigg( -10 \, t^{2} - \frac{5}{3} \, t^{3} \bigg) \mathbf{j} \), substitute \(t = 0.02\, \textrm{s}\): \[ y(t) = -10 (0.02)^{2} - \frac{5}{3} (0.02)^{3} \, \]. Hence, \[ y(0.02) = -10 (0.0004) - \frac{5}{3} (0.000008) \,= -0.004 - 0.0000133 = -0.0040133 \, \mathrm{m} = -4.0133 \, \mathrm{mm} \].

06

Calculate the angle the velocity vector forms with the horizontal when leaving the box (Part e)

First, find the components of the velocity function at \(t = 0.02 \, \textrm{s}\). From the velocity function, \( \textbf{v}(t)=100 \, \mathbf{i} + (-20 \, t - 5 \, t^{2}) \mathbf{j} \), substitute \(t = 0.02 \, \textrm{s}\): \[ v_{x} = 100 \, \mathrm{m/s} \, \]; \[ v_{y} = -20 (0.02) - 5 (0.02)^2 \, \]; Hence, \[ v_{y} = -0.4 - 0.002 = -0.402 \, \mathrm{m/s} \, \]. To find the angle \( \theta \), use \[ \theta = \arctan \bigg( \frac{v_{y}}{v_{x}} \bigg ) \]. \[ \theta = \arctan \bigg( \frac{-0.402 \, \mathrm{m/s}}{100 \, \mathrm{m/s}} \bigg ) = \arctan (-0.00402) \]. Therefore, \[ \theta \approx -0.230^{\circ} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electron Motion

Electrons are tiny particles with negative charge that are part of atoms. When they are allowed to move freely, they can be controlled and manipulated using electric fields. In this exercise, an electron is initially moving straight in the x-direction with a velocity of 100 meters per second (m/s). Once the electron enters a region with a varying electric field, its motion changes.

Understanding electron motion involves both its path and how its velocity and position change over time. The electric field applies a force on the electron, altering its trajectory. By analyzing the effects of this electric field, we can predict how the electron will behave.

The initial conditions of the problem set the electron in motion, and further calculations are needed to understand this dynamic movement.

Varying Electric Field

The electric field in this scenario changes with time and consequently affects the electron's acceleration. The given acceleration function is: \[ \textbf{a}(t) = \big(-20 \frac {m} {s^2} - 10 \frac {m} {s^3} t \big) \textbf{j}\]

Here, the electric field causes an initial constant acceleration of -20 m/s² and an additional time-dependent component of -10 m/s³ multiplied by time, causing it to vary as time progresses.

In simpler terms, as the electron spends time in the field, its acceleration becomes more negative in the y-direction. It means that the electron experiences a greater pull downward (in -j direction) as time goes on.

Displacement Calculation

To find where the electron will be at any given time, we need to calculate its displacement. Displacement tells us how far the electron has moved from its original position.

Since velocity is the rate of change of position, we integrate the velocity function to get the position as a function of time. Starting from the velocity equation:
\[ \textbf{v}(t) = 100 \frac{m}{s} \textbf{i} + (-20 t \frac {m} {s^2} - 5 t^2 \frac {m} {s^3}) \textbf{j} \]

By integrating this, we find:
\[ \textbf{r}(t) = 100 t \frac{m}{s} \textbf{i} + \big( -10 t^2 \frac {m} {s^2} - \frac{5}{3} t^3 \frac {m} {s^3} \big) \textbf{j} \]

This equation gives the position at any time 't', showing how far the electron has traveled both in the x- and y-directions.

Velocity Calculation

Calculating the electron's velocity involves understanding how the speed and direction change over time. The initial velocity is given as 100 m/s in the x-direction. The y-velocity, influenced by the varying electric field, is found by integrating the acceleration equation.

Starting from \[ \frac{d \textbf{v}(t)}{dt} = \textbf{a}(t) \]
We integrate the given acceleration function:
\[ \textbf{a}(t) = \big(-20 \frac {m} {s^2} - 10 \frac {m} {s^3} t \big) \textbf{j} \]
Integrating, we get:
\[ \textbf{v}(t) = 100 \frac{m}{s} \textbf{i} + (-20 t \frac {m} {s^2} - 5 t^2 \frac {m} {s^3}) \textbf{j} \]
This tells us that the x-component of the velocity remains the same, while the y-component depends on time, decreasing continuously as time increases due to the negative acceleration.

Acceleration Integration

Integration of acceleration is key to understanding how velocity and position change over time. Acceleration indicates the rate at which velocity changes. By integrating it once, we find the velocity; integrating again, we find the position.

The given acceleration function is:
\[ \textbf{a}(t) = \big(-20 \frac {m} {s^2} - 10 \frac {m} {s^3} t \big) \textbf{j} \]

Integrating with respect to time to get velocity:
\[ \textbf{v}(t) = 100 \frac{m}{s} \textbf{i} + \big( -20 t \frac {m} {s^2} - 5 t^2 \frac {m} {s^3} \big) \textbf{j} \]

Integrating the velocity gives the position:
\[ \textbf{r}(t) = 100 t \frac{m}{s} \textbf{i} + \big( -10 t^2 \frac {m} {s^2} - \frac{5}{3} t^3 \frac {m} {s^3} \big) \textbf{j} \]

Understanding these integrations helps explain how an electron's speed and position change under a varying electric field.