Magnitude of the charge on each plate

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SUMMARY

The discussion centers on calculating the magnitude of the charge on each plate of a capacitor, given a small plastic ball with a mass of 9.51 x 10^-3 kg and a charge of +0.206 micro Coulombs. The ball is suspended at an angle of 30.0 degrees from the vertical, indicating equilibrium between gravitational and electric forces. Key equations utilized include the electric field equation E = q/ε₀A and force balance equations. The solution involves determining tension in the thread and relating it to the electric force acting on the ball.

PREREQUISITES
  • Understanding of electric fields and forces, specifically E = q/ε₀A
  • Knowledge of equilibrium conditions in physics
  • Familiarity with free body diagrams (FBD)
  • Basic algebra for solving equations
NEXT STEPS
  • Study the concept of electric fields in capacitors, focusing on E = q/ε₀A
  • Learn about the equilibrium of forces in physics, particularly in charged systems
  • Explore the derivation and application of free body diagrams in various scenarios
  • Practice solving problems involving tension and electric forces in equilibrium situations
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Students studying physics, particularly those focusing on electromagnetism and mechanics, as well as educators looking for practical examples of charge interactions in capacitors.

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Homework Statement


A small plastic ball of mass 9.51 X 10^-3kg and charge of +0.206 micro Coulombs is suspended from an insulating thread and hangs between the plates of a capacitor. The ball is in equilibrium, with the thread making an angle of 30.0 degrees with respect to the vertical. The are of the plate is 0.01352m2. What is the magnitude of the charge on each plate.


Homework Equations



E=q/epsilon0A

Fe+FG+T=0

Fe - TSin30=0

TCOS30=mg

The Attempt at a Solution



So I drew a FBD and decided that since it was at equilbrium the force of attraction and repulsion of the plates was zero. And that the only other forces were mg and T. I am just having trouble understanding what to do with the tension and how I am supposed to relate that to the magnitude of charge on each plate.
 
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If you treat the plastic ball as a point charge, then the electric force is F = q_b*E (q_b is the charge on the ball)

Once you solve for T, you should be able to substitute your expression for the electric field into your equation for force in the horizontal direction and simply solve for q
 

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