Magnitude of the electric field between the plates

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SUMMARY

The discussion focuses on calculating the electric field magnitude between two parallel plates connected to a 13.0V battery, separated by 7.00x10-4 m. The electric field (E) is determined using the formula E = V/d, resulting in an electric field of approximately 18,571.43 V/m. Additionally, the change in electric potential energy (ΔU) for a charge of -4.29x10-6 C moving between the plates is calculated using ΔU = qV, yielding a value of -0.0000575 J.

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  • Understanding of electric fields and potential difference
  • Familiarity with the formula E = V/d
  • Knowledge of charge and electric potential energy calculations
  • Basic grasp of units such as volts (V), coulombs (C), and joules (J)
NEXT STEPS
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  • Learn about the relationship between electric potential and electric potential energy
  • Explore the effects of varying plate separation on electric field magnitude
  • Investigate the role of dielectric materials in electric fields
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A uniform electric field exists between two oppositely charged parallel plates connected to a 13.0V battery. The plates are separated by 7.00x10-4 m.a) Determine the magnitude of the electric field between the plates
b) If a charge of -4.29x10-6 C moves from one plate to another, calculate the change in electric potential energy of the charge.

attempt:

σ = Q / L² = 13.0 μC / (7.00x10^-4)²

E = σ / ε₀
?
 
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