Relationship: Force on a charge at some distance from capacitor Plates

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SUMMARY

The discussion centers on the forces experienced by a charge placed at different distances from two parallel capacitor plates with uniform charge distributions. The key equations involved are F=qE, where the force on a test charge is proportional to the electric field, and E = η/ε0, which describes the electric field between the plates. It is established that the electric field remains constant between the plates, leading to the conclusion that |F1| equals |F2|, despite the varying distances. The assumption that the test charge does not disturb the uniform charge distribution on the plates is critical to validating this conclusion.

PREREQUISITES
  • Understanding of electric fields and forces, specifically F=qE.
  • Knowledge of capacitor theory, particularly the behavior of parallel plate capacitors.
  • Familiarity with charge distribution concepts, including η = Q/A.
  • Basic grasp of electrostatics and the implications of charge placement in electric fields.
NEXT STEPS
  • Study the derivation and implications of the electric field equation E = η/ε0 in greater detail.
  • Explore the concept of superposition in electric fields and its effects on charge interactions.
  • Investigate the impact of varying charge magnitudes on force calculations in electrostatics.
  • Learn about the limitations of the assumption that test charges do not affect the electric field in practical scenarios.
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding electrostatics and the behavior of charges in electric fields, particularly in the context of capacitors.

Minhtran1092
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Homework Statement



Two capacitor plates (infinite plane) with uniform charge distribution but different charges (one positive, one negative) are distance "d" apart. A charge is place at d/2 and feels force |F1|. A charge is placed at d/4 and feels force |F2|. How does |F1| compare to |F2|?

Homework Equations



F=qE (Force on a test charge is proportional to the electric field at the point charge)
E = η/ε0 (η = Q/A, charge distribution); Electrical field between two parallel plate capacitors

The Attempt at a Solution



The electrical field between two parallel plate capacitors is constant and uniform in direction at any point. Since F=qE, E is constant and 'q' is negligible, F≈E everywhere. Thus, |F1|=|F2|.
 
Last edited:
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That looks good except for the statement F≈E. q is assumed negligible in the sense that it is assumed q does not disturb the uniform charge distributions on the plates. But that doesn't mean that you can ignore q in the equation F = qE. Do you see that your answer is still valid even for F = qE?
 
Last edited:

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