Calculate the electric field strength inside and outside a wire

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SUMMARY

The discussion focuses on calculating the electric field strength inside and outside an infinitely long conducting cylinder with radius R and uniform surface charge density (lambda per unit length). It is established that the electric field inside the conductor is zero due to the properties of conductors. For the electric field outside the cylinder, the correct formula to use is E = q/(4πε₀R²), where q is the total charge enclosed by the Gaussian surface. The area of the Gaussian surface must be considered as the lateral surface area of the cylinder, which is 2πRh.

PREREQUISITES
  • Understanding of Gauss's Law
  • Familiarity with electric field concepts
  • Knowledge of surface charge density
  • Basic calculus for integration
NEXT STEPS
  • Study Gauss's Law in detail
  • Learn about electric fields in conductors
  • Explore the concept of surface charge density
  • Practice problems involving cylindrical symmetry
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Students studying electromagnetism, physics educators, and anyone interested in understanding electric fields in conductive materials.

matt_crouch
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Homework Statement



An infinitly long conducting cylinder of Radius R carries a inform surface charge of (lambda per unit length) determine the electric field strength inside and outside the cylinder

Homework Equations



integral (E.ds)= q/e0

The Attempt at a Solution



im not really sure what to do at all i tried simply differentiating the equation above and substituting lambda in for q but I am pretty sure that's not right...
 
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First, the field inside is easiest to start with. The q in integral (E.ds) = q/e0, is the charge inside a gaussian surface. So a gaussian surface inside the conductor encloses how much charge? ... so the electric field is...?

Next, you need to simply integrate the left side of the equation after subtituting dr for ds. Then use a gaussian surface of some length l, and figure out what the q enclosed is and plug that into the right side.
 
so since its a Gaussian surface do i calculate the line charge density by integrating lambda from R to infinity? then substitute that into the equation above?
 
Electric field inside the cylinder is zero. Because that is conductor material
 
ok so outside the conductor do i just use the E=q/4(pi)e0R^2 ?
 
matt_crouch said:
so outside the conductor do i just use the E=q/4(pi)e0R^2 ?
It's not correct. What is area of Gaussian surface?
 
ahhh so i need to use the area of a cylinder? so 2(pi)rh +2(pi)r^2 ?
 

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