Question

Humid air brakes down (its molecules become ionized) in an electric field of$\mathbf{3}\mathbf{.0}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{\text{\hspace{0.17em}N/C}}$. In that field, what is the magnitude of the electrostatic force on (a) an electron and (b) an ion with a single electron missing?

Short answer

1. The magnitude of the electrostatic force on an electron is$4.8\times {10}^{-13}\text{\hspace{0.17em}N}$
2. The magnitude of the force on an ion with a single electron missing is$4.8\times {10}^{-13}\text{N}$

Step-by-step solution

The given data

• The given electric field,$E=3.0\times {10}^6\text{\hspace{0.17em}N/C}$

Understanding the concept of electric field

When a charged particle is placed in an electric field, it experiences a electrostatic force that is equal to the product of its charge and electric field.

Using the concept of force to the electric field, we can get the magnitude of force for the given cases.

Formula:

The magnitude of the electrostatic force on a point charge can be given by,

$F=qE$ (i)

where E is the magnitude of the electric field at the location of the particle.

a) Calculation of the electrostatic force on an electron

Using equation (i), the force on an electron is given as:

$\begin{array}{c}{F}_e=(1.60\times {10}^{-19}\text{\hspace{0.17em}C})(3.0\times {10}^6\text{\hspace{0.17em}N/C})\\ =4.8\times {10}^{-13}\text{\hspace{0.17em}N}\end{array}$

Hence, the value of the force is

b) Calculation of the force on an ion

Using the given data in equation (i), the electrostatic force on an ion having single electron missing is given as:

$\begin{array}{c}{F}_i=(1.60\times {10}^{-19}\text{\hspace{0.17em}C})(3.0\times {10}^6\text{\hspace{0.17em}N/C})\\ =4.8\times {10}^{-13}\text{\hspace{0.17em}N}\end{array}$

Hence, the value of the force is$4.8\times {10}^{-13}\text{\hspace{0.17em}N}$