Absolute potential from sphere.

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SUMMARY

The absolute potential at a distance of 3 meters from the center of a conducting sphere with a radius of 5 cm and a surface charge density of 1 pC/m² is calculated using the formula V=Q/4∏εr. The charge Q is determined by integrating the surface charge density over the sphere's surface area, yielding a total charge of approximately 0.0314159 pC. Substituting this charge into the potential formula with ε set to 2 times the permittivity of free space (8.85 x 10^-12 F/m) results in an absolute potential of 47.0809 µV, consistent with the answer provided in the 5th edition of "Electromagnetics with Applications."

PREREQUISITES
  • Understanding of electrostatics and electric potential
  • Familiarity with the concept of surface charge density
  • Knowledge of the permittivity of free space and its application
  • Ability to perform integration over a sphere's surface area
NEXT STEPS
  • Study the derivation of the electric potential from point charges
  • Learn about the effects of relative permittivity on electric fields
  • Explore the applications of Gauss's Law in electrostatics
  • Investigate the relationship between charge density and electric potential
USEFUL FOR

Students of electromagnetism, electrical engineers, and physicists interested in understanding electric potential and charge distributions in conductive materials.

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Homework Statement



A conducting sphere of 5-cm radius has a surface charge density of 1 pC/m^2. If the medium of relative permittivity is 2. How can we find the absolute potential at a distance of 3m from the center of the sphere?

Homework Equations



V=Q/4∏εr

The Attempt at a Solution


In this case, i make r=3, and then sub to formula. Is that right??
 
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The charge caused by the sphere, if you integrated over the surface of the sphere, would be 4*pi*r^2*1pC/m^2, where r=0.03m. Which equals pi/100 pC. Really, just the surface area of the sphere times the charge density.
Plugging this charge into the equation as "Q", with ε=2*8.85*10^-12, r=3m, we get:
(0.0314159)*10^-12/(3*4*pi*8.85*2*10^-12)= 47.0809*10^-6V or 47.0809uV. Which is the answer given in your book, assuming you're using the 5th edition of Electromagnetics with Applications.
 

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