V of 2 parallel lines makes cylinders

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SUMMARY

The discussion centers on calculating the electric potential V(x,y,z) for infinitely long wires with charge densities ±λ located at ±a on the y-axis. The solution for part A is straightforward, but part B requires demonstrating that the equipotential surfaces are circular cylinders. The key equations involved include Gauss's law and the gradient of the potential, specifically E = -∇V. The conclusion is that the equipotential surfaces are indeed circular cylinders, with the axis along the y-axis and a radius determined by the potential V0.

PREREQUISITES
  • Understanding of Gauss's law in electrostatics
  • Familiarity with electric potential and its relationship to electric fields
  • Knowledge of cylindrical coordinate systems
  • Ability to perform vector calculus operations, specifically gradients
NEXT STEPS
  • Study the application of Gauss's law in different geometries
  • Learn about electric potential and equipotential surfaces in electrostatics
  • Explore cylindrical coordinates and their applications in physics
  • Investigate the mathematical derivation of electric fields from potentials
USEFUL FOR

This discussion is beneficial for physics students, particularly those studying electromagnetism, as well as educators and anyone interested in the mathematical foundations of electric fields and potentials.

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Homework Statement


Expanded from Griffith's (3rd ed) #2.47

Infinitely long wires parallel to x-axis carrying ±λ charge densities intersect the y-axis at ±a.

A) Calculate V(x,y,z) if V(0,0,0)=0
B)Show that the equipotential surfaces are circular
cylinders. Locate the axis and calculate the
radius of the cylinder that is at potential V0.

Homework Equations


Gauss, -∫Edl
E=-∇V
?x2+y2=r2?

The Attempt at a Solution



I've found the solution to part A easily enough but I'm not sure how to approach B. I know that at the origin the potential is zero along the x-axis but I'm confused by how I should show that these are both circular cylinders.

I'd post my solution for A but it's exactly the same as the solution manual so it's easy enough to find.

Any help would be appreciated.
 
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I'd post my solution for A but it's exactly the same as the solution manual so it's easy enough to find.

Why not look at your solutions manual. . .
 
The part b is not part of the original question. Thus, not in the manual.
 
hmm, never mind I guess it was in the manual. Sorry thought that he just added it. thanks anyway.
 

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