Question

In Fig. 22-61, an electron is shot at an initial speed of,${\mathit{v}}_{\mathbf{0}}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.00}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{6}}\mathbf{}\mathbf{\text{m/s}}$ at angle$\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{40}\mathbf{.0}\mathbf{°}$from an xaxis. It moves through a uniform electric field. A screen for detecting electrons is positioned parallel to the yaxis, at distance$\mathit{x}\mathbf{}\mathbf{=}\mathbf{3}\mathbf{.00}\mathbf{}\mathbf{\text{m}}$. In unit-vector notation, what is the velocity of the electron when it hits the screen?

Short answer

The velocity of the electron when it hits the screen is.$(1.53\times {10}^6\text{m/s})\widehat{i}-(4.34\times {10}^5\text{m/s})\widehat{j}$

Step-by-step solution

The given data

1. Initial speed of electron,$v_0=2.00x{10}^6\text{m/s}$
2. The angle with x axis,${\theta }_0={40.0}^o$
3. The uniform electric field,$\overrightarrow{E}=(5.00\text{N/C})\widehat{j}$
4. Screen distance parallel to the x-axis,$x=3.00\text{m}$

Understanding the concept of electric field and kinematic equations

Since the electron is negatively charged, then (as a consequence of the magnitude of the electrostatic force on a point charge of magnitude q (given by F = qE, where E is the magnitude of the electric field at the location of the particle) with Newton’s second law) the field E pointing in the +y direction (which we will call “upward”) leads to a downward acceleration. This is exactly like a projectile motion problem (but with g replaced with, acceleration

$\begin{array}{c}\mathbf{a}\mathbf{=}\mathbf{e}\mathbf{E}\mathbf{/}\mathbf{m}\\ \mathbf{=}\mathbf{8}\mathbf{.78}\mathbf{x}{\mathbf{10}}^{\mathbf{11}}\mathbf{\text{m/s}}\end{array}$).

Formulae:

Using the concept of projectile motion, the time of travel by the electron,

$t=\frac{x}{v_{o}cos{\theta }_o}$(i)

Now, the velocity of the electron using projectile motion,$v_y=v_{o}sin{\theta }_o-a{t}^{}$ (ii)

Calculation of the velocity of the electron

$\begin{array}{c}t=\frac{3.00\text{m}}{(2.00\times {10}^6\text{m/s})cos{40.0}^o}\\ \cap =1.96\times {10}^{-6}\text{s}\end{array}$

The required velocity component using equation (ii) is given as:

$\begin{array}{c}{v}_y=(2.00\times {10}^6\text{m/s})sin{40.0}^o-(8.78\times {10}^{11}{\text{m/s}}^{\text{2}})(1.96\times {10}^{-6}\text{s})\\ =-4.34\times {10}^5\text{m/s}\end{array}$

Since the x component of velocity does not change, then the final velocity is.

$(1.53\times {10}^6\text{m/s})\widehat{i}-(4.34\times {10}^5\text{m/s})\widehat{j}$