Application of Gauss's Law to Charged Insulators

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SUMMARY

The discussion focuses on applying Gauss's Law to a cylindrical shell with a radius of 7.00 cm and a length of 240 cm, which has a uniform charge distribution on its curved surface. The electric field at a point 19.0 cm from the axis is measured at 36.0 kN/C. Participants clarify the correct application of the formula Flux = Q/e = E*2π*r*l, addressing the confusion regarding charge density and the correct value of the permittivity constant. The net charge on the shell is confirmed to be +913 nC after correcting the value of e used in calculations.

PREREQUISITES
  • Understanding of Gauss's Law and its application to cylindrical symmetry
  • Familiarity with electric field calculations and charge density concepts
  • Knowledge of the permittivity of free space (ε₀) and its role in electrostatics
  • Basic algebra skills for manipulating equations and solving for unknowns
NEXT STEPS
  • Study the derivation and applications of Gauss's Law in electrostatics
  • Learn about electric field calculations for different geometries, including spherical and planar symmetries
  • Explore the concept of charge density and its impact on electric fields
  • Review the significance of the permittivity constant (ε₀) in various electrostatic equations
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Students and educators in physics, particularly those focusing on electrostatics and electromagnetism, as well as anyone seeking to understand the application of Gauss's Law to charged insulators.

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Homework Statement



A cylindrical shell of radius 7.00 cm and length 240 cm has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 19.0 cm ra- dially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C. Use approximate relationships to find (a) the net charge on the shell and (b) the electric field at a point 4.00 cm from the axis, measured radially outward from the midpoint of the shell.

Homework Equations


The Attempt at a Solution



Could someone please explain to me why this does not work:

Flux = Q/e = E*2pi*r*l

Where I am thinking of Q in terms of an average AREA charge density * the area, and not a line density * a length.
-- Is it because it lacks enough independent equations?
 
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Why do you think that doesn't work?
 
Because I am currently getting a different answer than that given by the book - which is:

+913 nC
 
Okay it works... I've been using an incorrect value for e when solving from k - apologies.
 

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