Question

Your physics assignment is to figure out a way to use electricity to launch a small $6.0-cm$-long plastic drink stirrer. You decide that you’ll charge the little plastic rod by rubbing it with fur, then hold it near a long, charged wire, as shown in FIGURE P23.56. When you let go, the electric force of the wire on the plastic rod will shoot it away. Suppose you can uniformly charge the plastic stirrer to $10nC$and that the linear charge density of the long wire is $1.0\times {10}^{-7}C/m$. What is the net electric force on the plastic stirrer if the end closest to the wire is $2.0cm$away?

Hint: The stirrer cannot be modeled as a point charge; an integration is required.

Short answer

The net electric force on the plastic stirrer is$\mathbf{4}\mathbf{.}\mathbf{2}\mathbf{}\mathit{x}\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{}\mathit{N}$

Step-by-step solution

Given information and formula used 

Given :

Length of plastic drink stirrer :$6.0-cm$

The plastic stirrer is uniformly charged to : $10nC$

The linear charge density of the long wire : $1.0\times {10}^{-7}C/m$

The end closest to the wire : $2.0cm$away

Theory used :

The charge is

The electric field produced by a long charge wire is : $E=\frac{2\lambda }{4\pi {\epsilon }_{0}x}$

Evaluating the required net electric force 

Cut an infinitesimal segment $dn$off the lengthy charge wire at a distance of $n$. The charge is as follows:

$dn=q=\frac{Q}{L}dn$

At a distance of $n$, the electric field produced by a long charge wire is:

$E=\frac{2\lambda }{4\pi {\epsilon }_{0}x}$

Integrating the equation from $2.0cm\mathrm{to}8.0cm$given the condition :

$$\begin{gathered} d{F}_e=qE \\ =\frac{2\lambda }{4\pi {\epsilon }_0}\frac{Q}{2}\times \frac{dn}{n} \\ \Rightarrow F_e=\frac{2\lambda }{4\pi {\epsilon }_0}\frac{Q}{2}{\int }_2^8\frac{dn}{n} \end{gathered}$$

Solve the problem for $F_e$.

$$\begin{gathered} F_e=\frac{2\lambda }{4\pi {\epsilon }_0}\frac{Q}{2}\ln (8-2) \\ =\frac{2\times 1\times {10}^{-7}\times 9\times {10}^9\times 10\times {10}^{-9}}{6x{10}^{-2}}\times \ln \left(4\right) \\ =4.15x{10}^{-4}N \\ \approx \boxed{\mathbf{4}\mathbf{.}\mathbf{2}\mathbf{}\mathit{x}\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{}\mathit{N}} \end{gathered}$$