Force on a Charged Bead Near a Line Charge

AI Thread Summary
A 10-cm long glass rod charged to +50 nC exerts a force on a -5.2 nC plastic bead located 4.1 cm from its center. The discussion involves calculating the force using the equation dF = (k dq)/r², with a focus on integrating the contributions from the charged rod. Participants clarify that the integration should consider a distance element (dz) instead of a charge element (dq), emphasizing the line charge concept. The integration approach involves treating the rod symmetrically and doubling the result from one half. The conversation highlights the importance of correctly setting up the problem for accurate calculations.
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Homework Statement


A 10-cm long thin glass rod is uniformly charged to +50 nC. A small plastic bead, charged to - 5.2 nC, is 4.1 cm from the center of the rod. What is the magnitude of the force on the bead?

Homework Equations


dF = (k dq)/r^{2}

The Attempt at a Solution


I've gotten to this point:

dF sin \theta = k\lambda * (dx/(h^{2} + x^{2})^{3/2}) * (h/\sqrt{h^{2} + x^{2}})

and I understand fairly well how and why. But now I have no idea where to go from here, aside from the fact that I have to integrate at some point. I know the two halves of the rod are equivalent so I can integrate from 0 to 5 and double (right?).

What do I do?
 
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Hootenanny said:
Furthermore, your initial equation should involve a distance element (dz), rather than a charge element (dq).
Isn't he using the line charge concept ? I guess so : dq = pdz, where the p denotes the charge per unit of distance.


marlon
 
marlon said:
Isn't he using the line charge concept ? I guess so : dq = pdz, where the p denotes the charge per unit of distance.


marlon
Now you mention it, yes it does :redface:. I'll go and put another coffee on...
 
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