Charge that a battery transfers between two plates

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SUMMARY

The discussion centers on calculating the charge transferred by a 1.5 V AA battery connected to a parallel-plate capacitor with 6.00 cm diameter plates spaced 4 mm apart. The capacitance (C) was calculated using the formula C = Aκε/d, resulting in a value of 1.99 x 10-12 F. The charge (Q) was then computed using Q = V * C, but the initial calculation was incorrect due to the omission of the π factor in the area formula. Correcting this error is essential for accurate charge determination.

PREREQUISITES
  • Understanding of capacitance and its formula C = Aκε/d
  • Knowledge of electric potential and charge relationships (V = Q/C)
  • Familiarity with the geometry of parallel-plate capacitors
  • Basic proficiency in algebra and unit conversions
NEXT STEPS
  • Review the derivation of the area formula for circular plates, including the π factor
  • Learn about the impact of dielectric constants (κ) on capacitance
  • Explore practical applications of capacitors in electronic circuits
  • Study the relationship between voltage, charge, and capacitance in various configurations
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Students in physics or electrical engineering, educators teaching capacitor concepts, and anyone involved in circuit design or analysis.

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Homework Statement


An AA battery (1.5 V potential) is connected to a parallel-plate capacitor having 6.00cm-diameter plates spaced 4mm apart. How much charge does the battery move from one plate to the other?

Homework Equations


V=Q/C
C=A\kappa\epsilon/d

The Attempt at a Solution


Plugged in values:
C = (.03)2*(8.85*10-12)/.004 = 1.99*10-12
Q = 1.5V * 1.99*10-12f

Getting an incorrect answer for Q. Thoughts?
 
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You forgot a factor of \pi in the formula for area.
 

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