Calculating Electric Field in a Charged Tube Using Gauss's Law

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SUMMARY

The discussion focuses on calculating the electric field within a hollow charged tube using Gauss's Law. The participants clarify that the inner radius is 'a' and the outer radius is 'b', with a uniform charge density denoted by ρ. The correct application of Gauss's Law leads to the electric field formula: E = 2πkρ(r² - a²)/r, where 'r' is the radial distance from the center of the tube. The confusion arises from the assumption about the tube being closed, which is crucial for the application of Gauss's Law.

PREREQUISITES
  • Understanding of Gauss's Law in electrostatics
  • Familiarity with electric field calculations
  • Knowledge of charge density concepts
  • Basic geometry of cylindrical coordinates
NEXT STEPS
  • Study the derivation of Gauss's Law for different geometries
  • Explore electric field calculations for hollow cylinders
  • Learn about the implications of closed versus open surfaces in Gauss's Law
  • Investigate alternative methods for calculating electric fields in non-closed systems
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Physics students, electrical engineers, and anyone interested in electrostatics and electric field calculations in cylindrical geometries.

yevi
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A long tube charge with charged with uniform spatial density \rho.
The inner radius of the tube is: a
The outer radius of the tube is: b

Need to find the electric field in: a<r<b

My approach is Gauss:

E*S=4 \pi kq

The S is the Gaussean Surface it should be 2 \pi r^2 ??

and q should be \rho*(r^2-a^2)??
 
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The answer should be:
E=2 \pi k\rho\frac{r^2-a^2}{r}\hat{r}

So I did something wrong...
 
yevi said:
The answer should be:
E=2 \pi k\rho\frac{r^2-a^2}{r}\hat{r}

So I did something wrong...

Is the tube closed? because if it isn't then Gauss law can't be used.
 
what do you mean closed?
The tube is hollow...

If I can't use gauss, what should i use?
 
I still don't understand why I can't use gauss here.
 
Anyone? :)
 

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