Charged ring, integrate for electric potential

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SUMMARY

The discussion focuses on calculating the electric potential along the axis of a charged ring with inner radius R1 and outer radius R2, carrying a uniform surface charge density σ. Participants emphasize the necessity of integrating to account for the varying distances from the charge elements to the observation point on the x-axis. The correct approach involves treating the ring as a collection of infinitesimally thin disks and deriving the potential through integration, leading to the expression V = (σ/2ε₀) * ∫(r dr / √(x² + r²)) from R1 to R2.

PREREQUISITES
  • Understanding of electric potential and its mathematical representation
  • Familiarity with integration techniques in calculus
  • Knowledge of surface charge density and its implications
  • Basic concepts of electric fields and their relation to potential
NEXT STEPS
  • Study the derivation of electric potential for different charge distributions
  • Learn about the application of integration in electrostatics
  • Explore the concept of surface charge density in detail
  • Investigate the use of cylindrical coordinates in electric field calculations
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Students in physics, particularly those studying electromagnetism, as well as educators and anyone looking to deepen their understanding of electric potential calculations in relation to charged objects.

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Homework Statement



A flat ring of inner radius R1 and outer radius R2 carries a uniform surface charge density σ. Determine the electric potential at points along the axis (the x axis). [Hint: Try substituting variables.]

Homework Equations



V = (kQ)/r


The Attempt at a Solution



As you can see from my screenshot, I think I've figured it out mostly I'm just stuck on finding the value for r in the above equation. Shouldn't it be something like sqrt(x^2+y^2)? But that can't be the answer because there's no labelled y-axis...
 

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Well, you can't calculate the potential directly from the equation since not every point on the disk will be the same distance from a given point along the x axis. Try splitting the disk into thin disks, and then doing an integration.
 
What exactly do you mean by "thin disks"? Don't I still need to find the distance between the x point and any given slice of charge on the disk?
 
By thin I mean infinitely thin. With the disk you're using, the distance to the x point is going to vary as you go farther out from the center. You can find an expression for the distance in terms of the x point and the point you go radially outward, but it's going to be variable, so the solution is going to involve integration -- if you just look at an infinitesimally thin disk and find an expression for the potential along its axis, then you can consider the original disk as a collection of these disks, and integrate them all together.
 
I'm having trouble finding an expression for the distance between any given point on the ring and the x-point, since there doesn't seem to be any variable that defines the distance between the center of the ring and any given piece of charge on the ring.
 
Not explicitly given in the diagram, but that doesn't mean you can't create your own. You can indentify a ring of charge with a radius, or if you really want to identify any point, a radius and an angle.
 
Thanks, I was finally able to figure it out, integrating from R1 to R2:

\frac{σ}{2ε_{0}}*\int\frac{rdr}{\sqrt{x^2+r^2}}

:approve:
 

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