How Does a Charged Glass Ball Polarize a Neutral Metal Foil?

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SUMMARY

The discussion focuses on the polarization of a neutral metal foil by a positively charged glass ball. The electric field generated by the glass ball, quantified as Q/4πε₀r², induces charges on the foil, creating a parallel plate capacitor effect. The correct relationship for the charge on the upper surface of the foil is derived as q = QR²/4h², where Q is the charge of the ball, R is the radius of the foil, and h is the height of the ball above the foil. The user initially miscalculated the charge due to an error in the surface area of the foil.

PREREQUISITES
  • Understanding of electric fields and Coulomb's law
  • Knowledge of electrostatics and conductors in electrostatic equilibrium
  • Familiarity with parallel plate capacitor concepts
  • Basic mathematical skills for solving equations involving electric fields
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Students studying electrostatics, physics educators, and anyone interested in understanding the behavior of charged objects and their effects on neutral conductors.

bcjochim07
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Please help! Electric Field Problem due tomorrow

Homework Statement


We want to analyze how a charged object picks up a neutral piece of metal. A small circular disk of foil is lying flat on a table. The foil has radius R and thickness t. A glass ball w/ positive charge Q is at height h above the foil.

The ball's electric field polarizes the foil. The foil surfaces with charges +q and -q then act as the the plates of a parallel plate capacitor with separation t. But the foil is a conductor in electrostatic equilibrium so the electric field inside the foil must be zero. Write this condition as a mathematical statement and use it to find the charge on the upper surface of the foil.


Homework Equations





The Attempt at a Solution



I know that the electric field of this sphere is Q/4[tex]\pi[/tex][tex]\epsilon[/tex]or^2

Should the electric field from the ball equal the electric field from the top surface of the plate?

Ecapacitor = [tex]\eta[/tex]/[tex]\epsilon[/tex]o

in the case of the foil disk [tex]\eta[/tex]=q/4[tex]\pi[/tex]r2

so the electric field is then q/4[tex]\pi[/tex]R2[tex]\epsilon[/tex]o

If I set this equal to the electric field of the ball I get

q/4[tex]\pi[/tex]R2[tex]\epsilon[/tex]o=Q/4[tex]\pi[/tex][tex]\epsilon[/tex]or^2

q=QR2/h2

The answer is actually q=QR2/4h2

Could someone help me understand this better?
 
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ok I found my mistake in the surface area of the foil... now for Q=50nC R= 1.0 mm and t=.01mm I need to find how close the ball must be to lift the foil
 

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