Insulating Cylindrical Shell, Potential at edge

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SUMMARY

The discussion focuses on calculating the electric field and electric potential of a long cylindrical insulating shell with an inner radius of 1.37 m and an outer radius of 1.60 m, featuring a charge density of 2.70 x 10-9 C/m3. The electric field at a distance of 1.81 m from the axis is determined to be 5.76 x 101 N/C, while at 1.52 m, it is 4.35 x 101 N/C. The potential at the outer radius is calculated using the integral of the electric field, confirming that the potential at the axis is zero.

PREREQUISITES
  • Understanding of electric fields and potentials in cylindrical coordinates
  • Familiarity with Gauss's Law and its application to cylindrical symmetry
  • Knowledge of integral calculus for evaluating electric potential
  • Concept of charge density and its role in electric field calculations
NEXT STEPS
  • Study the application of Gauss's Law for cylindrical shells
  • Learn about the relationship between electric field and electric potential
  • Explore advanced integral calculus techniques for electric potential calculations
  • Investigate the effects of varying charge densities on electric fields
USEFUL FOR

Students in physics, electrical engineering majors, and anyone interested in electrostatics and electric field calculations related to cylindrical geometries.

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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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