Question

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?

Short answer

The magnitude of the initial acceleration of the electron is$2.1\times {10}^{17}\mathrm{m}/{\mathrm{s}}^2$

Step-by-step solution

The given data

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

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

Understanding the concept of the electric field and Newtonian acceleration

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)

Calculation of the value of the initial acceleration of the electron

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 $2.1\times {10}^{17}\mathrm{m}/\mathrm{s}$.