Electric Potential of Ring of Charge: Q, R, x

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SUMMARY

The electric potential at a distance x above the center of a ring of charge Q with radius R is given by the formula V = k_e Q / √(R² + x²). The electric field E_x derived from this potential is E_x = k_e Q x / (R² + x²)^(3/2). The discussion clarifies that the radius of the ring should be denoted as R instead of a, correcting a common mistake in notation. The expressions provided for both potential and electric field are confirmed to be accurate.

PREREQUISITES
  • Understanding of electric potential and electric fields
  • Familiarity with calculus, specifically differentiation
  • Knowledge of the constant k_e (Coulomb's constant)
  • Basic concepts of charge distribution
NEXT STEPS
  • Study the derivation of electric potential for different charge distributions
  • Learn about the applications of electric fields in electrostatics
  • Explore the concept of electric field lines and their significance
  • Investigate the relationship between electric potential and electric field in three-dimensional space
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Students studying electromagnetism, physics educators, and anyone interested in understanding the principles of electric potential and fields in relation to charge distributions.

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Homework Statement


What is the electric potential a distance x above the center of a ring of charge Q with a
radius R? Derive the electric field from this relationship.


Homework Equations



V=k_e[tex]\int[/tex]dq/r = k_e[tex]\int[/tex]dq/([tex]\sqrt{}[/tex]a^2+x^2)

The Attempt at a Solution



V= k_e/([tex]\sqrt{}[/tex]a^2+x^2)[tex]\int[/tex] dq=k_eQ/([tex]\sqrt{}[/tex]a^2+x^2)

E_x=-dV/dx= -k_eQ(d/dx)(a^2+x^2)^-1/2
= -k_e(-1/2)(a^2+x^2)^-3/2(2x)
E_x= k_ex/(a^2+x^2)^3/2 Q

Is this right? If not what is?
 
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Your expressions are correct although, strictly speaking, the given radius of the ring is R not a.
 
opps...I think I was looking at another problem when I wrote that. Thanks
 

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