An electron is released 9.0cmfrom a very long non-conducting rod with a uniform$\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{\mu C}\mathbf{/}\mathbf{m}$. What is the magnitude of the electron’s initial acceleration?

a) Initial distance between the rod and electron,r=0.09 m

b) Linear charge density,$\lambda =6.0\times {10}^{-6}\mathrm{C}/\mathrm{m}$

Using the concept of the electric field, we can get the electrostatic force by substituting the value of the field in the force-electric field relation. Now, using Newton's second law of motion, the acceleration can be calculated for an electron by using the value of the electrostatic force.

Formulae:

The electric field of a long rod,

$\mathrm{E}=\frac{\mathrm{\lambda }}{2{\mathrm{\epsilon }}_0\mathrm{\pi r}}$ (1)

The force value is due to Newton’s second law,

$\mathrm{F}=\mathrm{ma}$ (2)

The electrostatic force of a charged particle,

$\mathrm{F}=\mathrm{qE}$ (3)

Substituting the value of the electric field of equation (1) in equation (2) and then combining Newton’s second law of equation (2) with the definition of the electric field of equation (3), we get the initial acceleration of the electron as follows:

$$\begin{gathered} \mathrm{ma}=\frac{\mathrm{e\lambda }}{2{\mathrm{\pi \epsilon }}_0\mathrm{r}} \\ \mathrm{a}=\frac{\mathrm{e\lambda }}{2{\mathrm{\pi \epsilon }}_0\mathrm{rm}} \\ \mathrm{a}=\frac{2\times \left(1.6\times {10}^{-19}\mathrm{C}\right)\left(6.0\times {10}^{-6}\mathrm{C}/\mathrm{m}\right)}{4{\mathrm{\pi \epsilon }}_0\left(0.09\mathrm{m}\right)\left(9.1\times {10}^{-31}\mathrm{kg}\right)} \\ \mathrm{a}=2.1\times {10}^{17}\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of the acceleration is $2.1\times {10}^{17}\mathrm{m}/\mathrm{s}$.