Absolute potential from sphere.

AI Thread Summary
To find the absolute potential at a distance of 3m from the center of a conducting sphere with a radius of 5 cm and a surface charge density of 1 pC/m^2, the formula V=Q/4∏εr is used. The total charge on the sphere is calculated by integrating the surface charge density over the sphere's surface area, resulting in a charge of approximately 0.0314159 pC. Substituting this charge into the potential formula with ε set to 2 times the permittivity of free space yields a potential of approximately 47.08 µV. This calculation aligns with the answer provided in the 5th edition of "Electromagnetics with Applications." The methodology confirms the correct approach to solving for the absolute potential.
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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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