Electric Potential of a Uniformly Charged Sphere

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SUMMARY

The electric potential of a uniformly charged sphere is given by the formula \(\frac{1}{4\pi \epsilon_0}\frac{Qr^2}{r_0^3}\), where \(Q\) is the total charge and \(r_0\) is the radius of the sphere. The confusion arises from the integration of the electric field, which is \(\frac{Q}{4\pi r^2 \epsilon_0}\). The discrepancy in the derived formula \(\frac{1}{8\pi \epsilon_0}\frac{Qr^2}{r_0^3}\) occurs due to the incorrect treatment of the charge \(Q\) as a constant rather than a variable dependent on \(r\). Clarification is needed on whether the potential is being calculated inside or outside the sphere.

PREREQUISITES
  • Understanding of electric potential and electric fields
  • Familiarity with Gauss's Law
  • Knowledge of calculus, specifically integration techniques
  • Concept of uniformly charged spheres and their properties
NEXT STEPS
  • Study the derivation of electric potential using Gauss's Law
  • Learn about the differences between electric potential inside and outside a uniformly charged sphere
  • Explore integration techniques for calculating electric fields from charge distributions
  • Review the concept of charge density and its impact on electric potential
USEFUL FOR

Physics students, electrical engineers, and anyone studying electrostatics or electric fields in charged objects.

Typhon4ever
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My book gives the electric potential of a uniformly charged sphere as \frac{1}{4\pi \epsilon_0}\frac{Qr^2}{r_0^3}. I can't derive this forumula. What I get is \frac{1}{8\pi \epsilon_0}\frac{Qr^2}{r_0^3}. I can guess how the book did it which is by taking the equation for voltage of a single point charge V=\frac{1}{4\pi \epsilon_0}\frac{Q}{r} then subbing in for Q enclosed Q\frac{r^3}{r_o^3} which is the charge in terms of r. That would give you the book answer. I tried doing it a different way however which I'm not sure why it won't work. I used Va-Vb=\int _a^{b} E dl. Yet when I get to solving for the electric field I get \frac{Q}{4\pi r^2 \epsilon_0}. If I take Q out then I can solve for the book answer. However, Q is in terms of r which I wrote above so I can't take Q out. I must integrate it. That gives me the 1/8 instead of the 1/4. What am I doing wrong? Thanks!
 
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Do you have a uniformly charged sphere, or ball? The derivation suggests a ball (charge everywhere in the volume), not a sphere. Q is the total charge, and r_0 is the radius of the ball?
How do you get the electric field? It should not depend on r like that.
 
Are you looking for the potential inside the sphere or outside? In other words is r > r0 or is r < r0?
 

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