Solving Electrical Physics Problem: Find Electric Force on Point Charge

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My brother asked me this question and i completely have no idea how to do cause it is a very long time. Can anyone help, or give me an idea on how to do


A rectangular insulating sheet carrying a negative charge Q evenly distributed over its surface is formed into a cylinder, producing a thin walled hollow object of length L and radius R.

• We wish to find the electric force on a positive point charge q, located on the axis of the cylinder at one end.


a. Draw a sketch showing the tube, the victim charge q, and an appropriate coordinate system for describing the problem. In your sketch clearly show E at the location of q.

b. Choose and then clearly indicate the charge element dq you will use for the integration in your sketch.


c. Set up the integration. That is express F as a constant times an integral over a single variable.

Thank you
 
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welcome to pf!

hi vtran0703! welcome to pf! :wink:

it's a fairly straightforward integration problem …

find the charge on a small area dzdθ of the cylinder, find the force at E from that charge, and then integrate over the whole cylinder

show us how far you get, and where you're stuck, and then we'll know how to help! :smile:
 
It's been quite a while since I saw these problem. To be honest, I wasn't good with them back then either, but I'll try, any help would be great
 
Remember that
[tex]\mathbf{E} = \int \frac{\lambda(\mathbf{r'}) d\mathbf{l}}{|\mathbf{r}-\mathbf{r'}|^2}\hat{(\mathbf{r}-\mathbf{r'})}[/tex]
or alternatively
[tex]\mathbf{E} = \int \frac{\lambda(\mathbf{r'}) \mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3 }d\mathbf{l}[/tex]
where r is the vector from the origin to a point in space and r' is a vector from the origin to the charge. So you make all the these infinitesimally different vectors as you sweep along the charge edge, and sum them up. A good idea for an origin is one where you can exploit symmetry in the problem, what's the most symmetric origin? What is the electric field from there?
 
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