What is the electric field above a uniformly charged cylinder?

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The discussion focuses on calculating the electric field above a uniformly charged cylinder using the concept of integrating the electric field contributions from individual disks. The user derived the electric field for a flat disk and attempted to extend this to a cylinder by integrating over its height. However, it was clarified that the integration should only cover the height of the cylinder from -L to 0, as the charge density above the cylinder is zero. The final expression for the electric field was confirmed to be correct, emphasizing the importance of considering the charge distribution accurately. The key takeaway is that the electric field above the cylinder is influenced solely by the charge within the cylinder itself.
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ok for one of my problem sets i have come across a problem I am a little unsure of:

a uniformly charged cylinder of radius R, length L, and volume charge density rho is aligned along the z-axis from z=0 to z=-L. Find the electric field a distance D above the top of the cylinder(ie at z=D).[ Hint consider the cylinder as a stack of disks of thickness dz.]

ok now i already computed the charge for a flat disk and obtained

E= \frac{ \sigma*z}{2*\epsilon}*(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}})<br />

so now i was thinking that all a cylinder is is many of these disks with a thickness dz i can simply take the integral


\frac{\rho}{2*\epsilon}\int_{-L}^{D}(1- \frac{z}{\sqrt{z^2+R^2}})dz

which produces the answer


\frac{\rho}{2*\epsilon}*(L+\sqrt{L^2+R^2}+D-\sqrt{D^2+R^2})


does this seem right? if not it would be great if someone could point me to my error

thanks
 
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You only have to integrate over the cylinder height fom -L to 0.
 
even if the electic field at point D is well above the cylinder?
 
ya the charge is only on the cylincer.
 
Charge density above the cylinder is zero.
 
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