Electric field in parallel plate capacitor

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SUMMARY

The discussion revolves around calculating the force acting on a point charge within an air-filled parallel plate capacitor with parameters d = 5mm, A = 500cm², and charge Q = 10nC. The correct approach involves using the electric field (E) between the plates, derived from the surface charge density (σ) and the displacement field (D), rather than calculating capacitance. The force on the point charge (Q' = 0.02Q) is determined using the formula F = EQ', leading to the correct answer of 4.52 * 10-6 N.

PREREQUISITES
  • Understanding of electric fields in capacitors
  • Familiarity with Gauss' theorem
  • Knowledge of surface charge density (σ) and displacement field (D)
  • Basic principles of electrostatics and force calculations
NEXT STEPS
  • Study the derivation of electric fields in parallel plate capacitors
  • Learn about surface charge density and its applications
  • Explore Gauss' theorem and its use in electrostatics
  • Investigate the relationship between electric field strength and force on charges
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Students studying electromagnetism, physics educators, and anyone involved in electrical engineering or capacitor design.

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Homework Statement



The parameters of an air-filled parallel plate capacitor are: d = 5mm and A = 500cm2. The
charge of the electrodes are Q = 10nC and −Q. Inside of the capacitor in the middle point
there is a point charge of 0, 02Q. Find the magnitude of the force acting to the point charge!

Homework Equations



I'm guessing these formulas

C=Q/V = ε0 * A/d

V = E*d

The Attempt at a Solution



The answer is supposed to be 4.52 * 10-6 N

However I plug in the values and get 9.02 N.. Am I using the wrong formulas, or am I just calculating it wrong?
 
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You don't want to compute capacitance.

What did you get for the E field between the plates? You can use for example σ = D where σ = surface charge density and D = εE. Or use Gauss' theorem.

Then F = EQ' where Q' = 0.02Q.

The given answer is correct BTW.
 

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