Solving Electric Potential: Find Charge on Parallel-Plate Capacitor

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SUMMARY

The charge on each plate of a parallel-plate capacitor can be determined using the formula Q = C * V, where Q is the charge, C is the capacitance, and V is the voltage. For a parallel-plate capacitor with an area of 0.0061 m² and a separation of 0.65 mm, the capacitance can be calculated using the formula C = ε₀ * (A/d), where ε₀ is the permittivity of free space (approximately 8.85 x 10⁻¹² F/m). When connected to a 12 V battery, the charge on each plate is found to be approximately 6.1 x 10⁻⁶ C.

PREREQUISITES
  • Understanding of capacitance and its formulas
  • Familiarity with the concept of electric potential
  • Knowledge of the permittivity of free space (ε₀)
  • Basic algebra for manipulating equations
NEXT STEPS
  • Learn how to calculate capacitance for different capacitor configurations
  • Study the relationship between charge, voltage, and capacitance in capacitors
  • Explore the effects of dielectric materials on capacitance
  • Investigate the applications of capacitors in electrical circuits
USEFUL FOR

Students in physics or electrical engineering, educators teaching capacitor concepts, and anyone interested in understanding the fundamentals of electric potential and capacitance.

spark2flame
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1. A parallel-plate capacitor filled with air has plates of area 0.0061 m2 and a separation of 0.65 mm. Find the magnitude of the charge on each plate when the capacitor is connected to a 12 V battery.



2. E = delta(V)/delta(s)



3. I keep getting the wrong answer by using the formula above, but I don't know how to solve for specific charges on plate capacitors!
 
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How does charge relate to capacitance and voltage? How do you find the capacitance of a parallel plate capacitor? (Look it up!)
 

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