Magnitude of the electric field between the plates

AI Thread Summary
The discussion focuses on calculating the magnitude of the electric field between two parallel plates connected to a 13.0V battery, separated by 7.00x10^-4 m. The electric field (E) can be determined using the formula E = V/d, resulting in an electric field of approximately 18,571 V/m. Additionally, the change in electric potential energy for a charge of -4.29x10^-6 C moving between the plates is calculated using the formula ΔU = qV, yielding a change of approximately -0.0000791 J. The calculations highlight the relationship between voltage, distance, and electric field strength in a uniform electric field. Understanding these principles is crucial for solving related physics problems.
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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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