Total Excess Charge using a Coulomb Balance

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SUMMARY

The discussion focuses on estimating the total excess charge on a conducting sphere charged to a potential of 2,000,000V using a Coulomb Balance experiment. The relevant equation for this calculation is V = q/R, where V is the potential, q is the charge, and R is the radius of the sphere. Additionally, the capacitance of a sphere is given by C = 4πε₀r, which simplifies calculations due to the symmetry of the sphere. Spheres are preferred in such experiments because they simplify the mathematical modeling of electric fields.

PREREQUISITES
  • Understanding of electrostatics and electric potential
  • Familiarity with Coulomb's Law
  • Knowledge of capacitance and its formulas
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the derivation of the capacitance formula C = 4πε₀r for spherical conductors
  • Explore the implications of electric potential in electrostatics
  • Learn about the applications of Coulomb Balance in experimental physics
  • Investigate the differences in electric field calculations for various geometrical shapes
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Physics students, educators, and researchers interested in electrostatics, particularly those conducting experiments involving charged conductors and electric fields.

ZedCar
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Homework Statement



If conducting a Coulomb Balance experiment, and a question asks to estimate the total excess charge on a sphere with a conducting surface charged to 2,000,000V potential.

Is there a specific equation I should be using to estimate this?

Thank you.

P.S. why are spheres used in such an experiment, and not some other shape?
 
Last edited:
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ZedCar said:

Homework Statement



If conducting a Coulomb Balance experiment, and a question asks to estimate the total excess charge on a sphere with a conducting surface charged to 2,000,000V potential.

Is there a specific equation I should be using to estimate this?

Thank you.

P.S. why are spheres used in such an experiment, and not some other shape?

V = q/R
'cause they make the math easy.
 
Is it not C = 4∏ε0r

(Sorry for large font, I was trying to obtain a small 0 beside ε!)
 
ZedCar said:
Is it not C = 4∏ε0r

(Sorry for large font, I was trying to obtain a small 0 beside ε!)


Yes, sure. In either case though you have to know the value of r.

(I assume you got the right formula for C of a sphere.)
 
Last edited:

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