Maximum potential (coaxial cable)

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SUMMARY

The maximum potential in a coaxial cable configuration is achieved when the inner conductor radius \( a \) is set to \( b/e \), where \( b \) is the outer conductor radius and \( e \) is the base of the natural logarithm. Utilizing Gauss's Law, the electric field \( E \) for the region between the conductors is expressed as \( E = \frac{2k\lambda}{r} \), with \( \lambda \) representing the charge per unit length \( Q/L \). The potential difference \( V \) is derived from the integral of the electric field, resulting in \( V = 2k\lambda \ln(b/a) \). To find the maximum potential, one must differentiate \( V \) with respect to \( a \) and set the derivative to zero.

PREREQUISITES
  • Understanding of Gauss's Law in electrostatics
  • Familiarity with electric field and potential difference concepts
  • Knowledge of logarithmic functions and their properties
  • Basic calculus, specifically differentiation techniques
NEXT STEPS
  • Study the application of Gauss's Law in cylindrical coordinates
  • Learn about electric field calculations in coaxial cable systems
  • Explore differentiation techniques for optimizing functions
  • Investigate the physical implications of potential differences in electrical engineering
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Students and professionals in electrical engineering, physicists studying electromagnetism, and anyone interested in optimizing coaxial cable designs for maximum potential.

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Homework Statement


Given a coaxial cable with outer conductor of radius b, and inner conductor of radius a. Prove that maximum potential is obtained when a = b /e, e is base of natural log.

Homework Equations


gauss law, ...

The Attempt at a Solution


By using Gauss Law,
for a < r < b,
i have E = 2*k*\lambda / r, where \lambda = Q/L

then the potential difference, V= \int E dr
V = 2*k*\lambda ln (b/a),

but i am not sure how to continue by differentiating V wrt a.
Help me pls...
Thank you.
 
Last edited:
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