Electric Potential on z-axis due to circular disk

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SUMMARY

The discussion centers on calculating the electric potential (V) on the z-axis due to a circular disk located in the xy-plane with radius b and uniform charge density (ρ_S). The correct formula for V is derived as V = (ρ_S / 2ε)[(z² + b²)^(1/2) - |z|]. The confusion arises from the term |z|, which is essential for accurately evaluating the potential at different positions along the z-axis. Participants emphasize the importance of correctly applying limits in the integral evaluation to arrive at the textbook's result.

PREREQUISITES
  • Understanding of electric potential and charge density concepts
  • Familiarity with calculus, specifically double integrals
  • Knowledge of electrostatics, particularly related to circular charge distributions
  • Proficiency in evaluating limits in integral calculus
NEXT STEPS
  • Study the derivation of electric potential from continuous charge distributions
  • Learn about the application of double integrals in electrostatics
  • Explore the significance of absolute values in mathematical expressions
  • Investigate the effects of varying charge densities on electric potential
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Students studying electromagnetism, physics educators, and anyone interested in understanding the mathematical foundations of electric potential due to charged objects.

sandy.bridge
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Homework Statement


Hey guys, I am a little bit confused how my textbook arrived at the answer it did for this particular question. The disk is in the xy-plane with radius b and uniform charge density. They go

V=\frac{\rho_S}{4πε}\int_0 ^{2π}\int_0 ^b\frac{r'}{(r'^2+z^2)^{1/2}}dr'd\phi'=\frac{ρ_S}{2ε}[(z^2+b^2)^{1/2}-|z|]

Where does the|z| come from? When I evaluate the integral, I do not get that.
I get
\frac{ρ_S}{2ε}[(z^2+b^2)^{1/2}]
 
Last edited:
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sandy.bridge said:

Homework Statement


Hey guys, I am a little bit confused how my textbook arrived at the answer it did for this particular question. The disk is in the xy-plane with radius b and uniform charge density. They go

V=\frac{\rho_S}{4πε}\int_0 ^{2π}\int_0 ^b\frac{r'}{(r'^2+z^2)^{1/2}}dr'd\phi'=\frac{ρ_S}{2ε}[(z^2+b^2)^{1/2}-|z|]

Where does the|z| come from? When I evaluate the integral, I do not get that.
I get
\frac{ρ_S}{2ε}[(z^2+b^2)^{1/2}]
Try to evaluate your limits again.

Here is a hint: Evaluating,

\sqrt{z^2 + r^2} \ \ \bigg|_{r=0}^b
is not just \sqrt{z^2 + b^2}. :wink:
 
! Thank you!
 

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