Electric Field Integration Outside a Cylinder

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SUMMARY

The discussion focuses on calculating the electric field generated by a thin, hollow cylinder with a charge of -90nC, a radius of 0.5 cm, and a length of 2.0 cm, specifically at a point 5.0 cm away from one end of the cylinder. The relevant equation for the electric field of a ring is provided as E=1/(4πε₀) * qΔz/(R²+Δz²), where R is the cylinder's radius and Δz is the distance from the ring to the observation point. The user is advised to represent the cylinder as a series of infinitesimal rings and to express the charge on each ring in terms of dz to facilitate integration.

PREREQUISITES
  • Understanding of electric fields and Coulomb's law
  • Familiarity with calculus, specifically integration techniques
  • Knowledge of the concept of charge distribution in physics
  • Basic understanding of the properties of alpha particles and their charge
NEXT STEPS
  • Study the derivation of the electric field from continuous charge distributions
  • Learn about the integration of electric fields from multiple point charges
  • Explore the concept of surface charge density and its application in cylindrical coordinates
  • Review the properties of alpha particles, including their charge and mass
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Students in physics, particularly those studying electromagnetism, as well as educators and anyone interested in the application of calculus to physical problems involving electric fields.

frostmephit
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Homework Statement


A thin, hollow cylinder missing its two end caps is shown to have a charge of -90nC. It has a radius of .5 cm and is 2.0 cm long. Find the charge on an [tex]\alpha[/tex] particle 5.0cm away from the end of the cylinder closest to the particle. (The center of the cylinder is on the same axis as the alpha particle.)


Homework Equations


(Electric Field of a Ring)
E=1/(4[tex]\pi\epsilon[/tex][tex]\o[/tex]) * q[tex]\Delta[/tex]z/(R^2+[tex]\Delta[/tex]z^2), where R is the radius of the cylinder and z is the distance from the center of the ring to the observation location, q is the charge, and the charge of an alpha particle is given by 2e


The Attempt at a Solution


The way I have gone about doing this is in such a manner as to divide the cylinder up into so many small rings, each with an infinitesimaly small charge. The problem is, I can't figure out how to create an equation representing the whole cylinder. I am aware that the charge on the rings would be equal to the surface area of one of them divided by the surface area of the greater cylinder times the charge, but I've been having trouble representing this as an equation for integration. Am I going in the right direction, and could someone please help?
 
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frostmephit said:

Homework Equations


(Electric Field of a Ring)
E=1/(4[tex]\pi\epsilon[/tex][tex]\o[/tex]) * q[tex]\Delta[/tex]z/(R^2+[tex]\Delta[/tex]z^2), where R is the radius of the cylinder and z is the distance from the center of the ring to the observation location, q is the charge, and the charge of an alpha particle is given by 2e


The Attempt at a Solution


The way I have gone about doing this is in such a manner as to divide the cylinder up into so many small rings, each with an infinitesimaly small charge. The problem is, I can't figure out how to create an equation representing the whole cylinder. I am aware that the charge on the rings would be equal to the surface area of one of them divided by the surface area of the greater cylinder times the charge, but I've been having trouble representing this as an equation for integration. Am I going in the right direction, and could someone please help?
Okay. First, I'll recommend using z instead of Δz, for the distance from the alpha particle to the ring.

If each ring has a "length" of dz along the z-direction, what would be the charge q on that ring? You're expression for q should contain a "dz" in it, so that will (hopefully) suggest an integral.
 

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