Q43P - Fundamentals Of Physics

Question

An electron is released from rest in a uniform electric field of magnitude$\mathbf{2}\mathbf{.00}\mathbf{ }\mathbf{\times }\mathbf{ }{\mathbf{10}}^{\mathbf{4}}\mathbf{\text{\hspace{0.17em}N/C}}$. Calculate the acceleration of the electron. (Ignore gravitation.)

Step-by-Step Solution

Answer

The acceleration of the electron is.$3.51\times {10}^{15}{\text{ m/s}}^{\text{2}}$

Step 1: The given data

Magnitude of electric field,$E=2\times {10}^4\text{\hspace{0.17em}N/C}$

Step 2: Understanding the concept of electric field

Newton’s second law states that the force on the particle is equal to the product of the mass and the acceleration of the particle. If the particle has a charge and it is placed in an electric field, it is acted upon by a electrostatic force that equal to the product of its charge and electric field.

 

Using the concept of Newton's law and electric field, we can calculate the acceleration of the electron in the given field.

 

Formulae:

 

The force due to Newton’s second law,$F=ma$ (i)

The force relation to the electric field of a particle, $F=qE$ (ii)

Step 3: Calculation of the acceleration of the electron

Using equation (ii) in equation (i) with the given data, we can get the acceleration of the electron as given:

$\begin{array}{c}a=\frac{eE}{m}\\ =\frac{(1.6 \times {10}^{-19}\text{C})\left(2.00 \times {10}^4\frac{\text{N}}{\text{C}}\right)}{9.11\times {10}^{-31}\text{ kg}}\\ =3.51\times {10}^{15}{\text{ m/s}}^{\text{2}}\end{array}$

Hence, the value of the acceleration is $3.51\times {10}^{15}{\text{ m/s}}^{\text{2}}$