Insulating Cylindrical Shell, Potential at edge

AI Thread Summary
The discussion revolves around calculating the electric field and potential for a cylindrical insulating shell with specified inner and outer radii and a uniform charge density. The electric field at a distance of 1.81 m from the axis is determined to be 5.76×10^1 N/C, while at 1.52 m, it is 4.35×10^1 N/C. The potential at the outer radius of the shell is calculated using the integral of the electric field, with the potential at the axis set to zero. The final equation used for potential incorporates the charge density, inner radius, and the differential radius element. The user successfully resolved the problem by applying these principles.
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Homework Statement



A long cylindrical insulating shell has an inner radius of a = 1.37 m and an outer radius of b = 1.60 m. The shell has a constant charge density of 2.70 10−9 C/m3. The picture shows an end-on cross-section of the cylindrical shell.

What is the magnitude of the electric field at a distance of r= 1.81 m from the axis?
5.76×101 N/C


What is the magnitude of the electric field at a distance of r= 1.52 m from the axis?
4.35×101 N/C

If we take the potential at the axis to be zero, what is the electric potential at the outer radius of the shell?


Homework Equations


V=kdq/r

Va - Vb=∫E.da


The Attempt at a Solution


∫ Q*(R^2 - r^2) / 2r *ε0 from the inner radius to the outer radius, my assumption was that since V equals the summation of electric field over the area enclosed, and that since potential was zero at the center, the potential would be equal to the difference in potential due to the electric field. Any insight or help is appreciated.

 
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hey for those interested I actually sorted myself out on this one, the final equation i used was the integral of rho(r^2 -a^2)*dr / 2 *r* e0 [a,R]. Where rho is the charge density, a is the interior radius, r is a variable, dr is the differential radius element, and e0 is epsilon naught, 8.854e-12.
 
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