Question

A uniform 500 N/C electric field points in the positive y-direction and acts on an electron initially at rest. After the electron has moved4.00 cm in the field, what is the energy of electron in eV?

Answer 1

Given data:

* The electric field in the y-direction is,

$E=500\text{ N/C}$

* The distance traveled by the electron is,

$\begin{gathered} d=4\text{ cm} \\ d=0.04\text{ m} \end{gathered}$

Solution:

The work done in terms of electric field is,

$W=\text{Edq}$

where q is the charge on an electron,

Substituting the known values,

$\begin{gathered} W=500\times0.04\times1.6\times10^{-19}\text{ J} \\ W=\frac{500\times0.04\times1.6\times10^{-19}}{1.6\times10^{-19}}\text{ eV} \\ W=500\times0.04\text{ eV} \\ W=20\text{ eV} \end{gathered}$

This work done is stored in the charge in form of energy.

Thus, the energy of the electron in eV is 20 eV.