Solving Physics Assignment: Launching a Plastic Stirrer with Electricity

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Homework Statement

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. When you let go, the electric force of the wire on the plastic rod will shoot it away. Suppose you can charge the plastic stirrer to 11.0 nC and that the linear charge density of the long wire is 1.0 *10^{ - 7C/m. What is the electric force on the plastic stirrer if the end closest to the wire is 2.0 cm away?

Homework Equations

force lawF=qE
Gauss's law \Phi_E = \int E*dA

The Attempt at a Solution

Well before I can solve this, I need to know the electric field that is coming off of the wire. So I use Gauss's law to determine that. I can place a cylinder around the wire and find the surface area of the cylinder excluding the ends. So

\Phi_E = \int E*dA = 2(pi)RLE=Q/\epsilon_{o}

Where 2(pi)RL is just the surface area of the wire. Moving the surface area to the other side you can solve for E.


E=Q/(\epsilon_{o}*2(pi)RL)

Since we know the charge density of the wire. we can replace Q with L\lambda since they equal each other. Plugging that in and simplifying gives you the new formula

E=\lambda/(\epsilon_{o}*2(pi)R)

And all you need to know is the radius of the charges away from the wire. My problem is how do you figure out where to take this radius from? I don't think the stirrer can be treated like a point charge. Where do I go from here?

 

[ideasrule]

Integration. I don't think you have any other choice.

 

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Integration. I don't think you have any other choice.

So wait. Should I take the charge density of the stirrer multiplied by dL (\lambda dL) and then replace R with (.02 +L)? Does that make any sense?

 

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I did out the problem and was able to get 4.57E-4 Newtons. Very happy that I was able to get this answer! I usually don't do integration like this to solve physics problems :).

Thanks for the help