Maximizing Electric Potential on a Uniformly Charged Cone

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SUMMARY

The discussion focuses on maximizing electric potential for a uniformly charged cone using the equation V = q/(4πE₀r). The participant struggles to find relevant information in their textbook and compares the cone's behavior to that of a uniformly charged sphere. They conclude that the potential can be derived using the differential equation dV = dq/(4πE₀r), emphasizing the importance of identifying turning points where the derivative is zero to locate maximum and minimum potentials.

PREREQUISITES
  • Understanding of electric potential and charge distributions
  • Familiarity with calculus, specifically derivatives and turning points
  • Knowledge of electrostatics, including Coulomb's law
  • Experience with differential equations in physics
NEXT STEPS
  • Study the properties of electric potential for different charge distributions, including cones and spheres
  • Learn about the method of finding turning points in functions
  • Explore the application of Gauss's law in electrostatics
  • Investigate numerical methods for solving differential equations in electrostatics
USEFUL FOR

Physics students, electrical engineers, and anyone studying electrostatics or electric potential in charged systems.

patm95
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Homework Statement



I am basically given a cone with uniform volume charge and told to find the area of highest electric potential.

Homework Equations



I want to use the equation: V= q/4piEr

The Attempt at a Solution



I am having trouble finding anything in my book. I am comparing this in my mind to a sphere and wondering if it would be prudent to find the center of mass of the cone and place a point as close to that as possible. I am just unsure if a cone of uniform volume charge is similar to a sphere of uniform volume charge in that it acts like a point charge outside of the volume. I can't seem to find any information on this.
 
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You'll want to use dV= dq/(4*pi*E_0*r) so that you can get the potential. Max and mins occur at turning points, when the derivative (or gradient in this case) is zero.
 
Ah. That makes a lot of sense. Thank you!
 

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