Q62P - Fundamentals Of Physics

Question

(a) what is the magnitude of an electron’s acceleration in a uniform electric field of magnitude $\mathbf{1}\mathbf{.40}\mathbf{ }\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{\text{N/C}}$? (b) How long would the electron take, starting from rest, to attain one-tenth the speed of light? (c) How far would it travel in that time?

Step-by-Step Solution

Answer

a) The magnitude of the acceleration of the electron is .$2.46\times {10}^{17}{\text{m/s}}^{\text{2}}$

b) The electron takes$1.22\times {10}^{-10}\text{s}$ long to attain one-tenth the speed of light.

c) It travels$1.83\times {10}^{-3}\text{m}$ far in that time.

Step 1: The given data

Magnitude of the electric field,$E=1.40x{10}^6\text{N/C}$

Step 2: Understanding the concept of force and acceleration

The acceleration of the electron is given by Newton’s second law:,$\mathit{F}\mathbf{=}\mathit{m}\mathit{a}$ where F is the electrostatic force.The magnitude of the force acting on the electron is,$\mathit{F}\mathbf{=}\mathit{q}\mathit{E}$ , where E is the magnitude of the electric field at its location.We then use the concepts of kinematical equations.

 

Formulae:

Using Newton’s second law, the acceleration of the electron,  $a=\frac{eE}{m}$        (i)

The first equation of kinematic motion,      $v-v_0=at$                                    (ii)

The third equation of kinematic motion,     $v^2-v_0^2=2ax$                  (iii)                

Step 3: a) Calculations of the magnitude of the acceleration

Using equation (i), we get the magnitude of the acceleration of the electron as:

$\begin{array}{c}a=\frac{(\text{1.60}\times {\text{10}}^{\text{-19}}\text{C})}{(\text{9.11}\times {\text{10}}^{\text{-31}}\text{kg})}(1.40\times {10}^6\text{N/C})\\ = 2.46\times {10}^{17}{\text{m/s}}^{\text{2}}\end{array}$

Hence, the value of the acceleration is.$2.46\times {10}^{17}{\text{m/s}}^{\text{2}}$

Step 4: b) Calculation of the time taken by the electron

With the velocity of the electron given as:

$\begin{array}{c}v=c/10\\ =3.00 x{10}^7\text{m/s}\end{array}$

The time taken by the electron using equation (ii) is given as:

1. $\begin{array}{c}t= \frac{(3.00\times {10}^7\text{m/s}-0\text{m/s})}{ (2.46\times {10}^{17}{\text{m/s}}^{\text{2}})}\\ =1.22\times {10}^{-10}\text{s}\end{array}$.

Hence, the value of the time taken is.$1.22\times {10}^{-10}\text{s}$

 

Step 5: c) Calculation of the distance travelled by the electron

Using kinematical equations we get the displacement using equation (iii) as follows:

$\begin{array}{c}\Delta x= \frac{{(3.00\times {10}^7\text{m/s})}^2-{\left(0\text{m/s}\right)}^2}{2 (2.46\times {10}^{17}{\text{m/s}}^{\text{2}})}\\ =1.83\times {10}^{-3}\text{m}\end{array}$

Hence, the distance travelled by the electron is.$1.83\times {10}^{-3}\text{m}$