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Please help! Electric Field Problem due tomorrow
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.
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?
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?