What is the surface charge density sigma?

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SUMMARY

The surface charge density (σ) of a 30.6-cm-diameter conducting sphere charged to 538 V is calculated to be 3.11E-08 C/m². The relationship between electric field (E), surface charge density (σ), and the permittivity of free space (ε₀) is defined by the equation E = σ / ε₀. To find the distance (r) at which the potential is 11.0 V, the equation r = (V/σ) * ε₀ is utilized, but the user encounters a discrepancy in the expected value of r.

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  • Understanding of electric potential and electric fields
  • Familiarity with the concepts of surface charge density and permittivity of free space
  • Knowledge of basic electrostatics and conducting spheres
  • Ability to manipulate equations involving electric potential and charge
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  • Learn about the concept of total charge on a conducting sphere
  • Explore the implications of charge distribution on electric fields
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Students studying electrostatics, physics educators, and anyone interested in understanding electric potential and charge distribution in conducting spheres.

dgoudie
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[SOLVED] Electric Potential

Homework Statement


20. [1pt]
A 30.6- cm-diameter conducting sphere is charged to 538 V relative to V = 0 at r =infinity. What is the surface charge density sigma?

Correct, computer gets: 3.11E-08 C/m^2 = Sigma, r=.153

21. [1pt]
At what distance will the potential due to the sphere be only 11.0 V?

Homework Equations


[tex]E=\sigma / \epsilon_{0} and<br /> <br /> E=V/r[/tex]

The Attempt at a Solution


I already found sigma for Question number 20, and for 21 I am trying the same method, but am just solving for r.

so I got [tex]r=(V/\sigma) *\epsilon_{0}[/tex]
and everytime I do it I come out with r=0.00313 m which is smaller than the initial r, even though I know from this statement "538 V relative to V = 0 at r =infinity" that if V is smaller r must be bigger.

What am I missing? Thanks in advance
 
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dgoudie said:
Correct, computer gets: 3.11E-08 C/m^2 = Sigma, r=.153

21. [1pt]
At what distance will the potential due to the sphere be only 11.0 V?

I'm not very sure what you mean by saying "computer gets". Do you know the concept you're applying?

The next question has nothing to do with sigma, but to do with the total charge. Find that.
 

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