5. Energy of a Capacitor in the Presence of a Dielectric

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SUMMARY

The discussion focuses on calculating the energy stored in a dielectric-filled parallel-plate capacitor with a plate area of 25.0 cm², a plate separation of 9.00 mm, and a voltage of 7.50 V. The dielectric constant is 2.00, and the relevant equations used include U=1/2C*V² and C=εA/d. The capacitance calculated is 2.07×10⁻¹⁰ F, leading to a stored energy of 8.59×10⁻⁶ J. The calculations utilize the permittivity constant of 8.85×10⁻¹² C²/N·m².

PREREQUISITES
  • Understanding of capacitor fundamentals, including capacitance and energy storage.
  • Familiarity with dielectric materials and their impact on capacitance.
  • Knowledge of the formula for capacitance: C=εA/d.
  • Ability to perform calculations involving electrical units such as volts, farads, and joules.
NEXT STEPS
  • Explore the effects of varying dielectric constants on capacitor performance.
  • Learn about different capacitor configurations and their applications in circuits.
  • Investigate the relationship between voltage, capacitance, and energy in capacitors.
  • Study advanced topics in electrostatics, including energy density in electric fields.
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Students and professionals in electrical engineering, physics enthusiasts, and anyone studying capacitor behavior in circuits, particularly in the context of dielectrics.

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Homework Statement


A dielectric-filled parallel-plate capacitor has plate area of 25.0 cm^2 and plate separation of 9.00 mm. The capacitor is connected to a battery that creates a constant voltage of 7.50 V. The dielectric constant is 2.00. Throughout the problem, use 8.85×10−12 C^2/N \cdot m^2


Homework Equations


U=1/2C*V^2
C=[tex]\epsilon[/tex]A/d
C=K*C_0



The Attempt at a Solution



2.07×10^−10
Keep in mind that the area of each portion of the capacitor is half that of the original capacitor.
 
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C=\epsilonA/dC=(2)(8.85 × 10^−12) (25.0 \ cm^2)/(0.009 \ m)=2.07×10^−10 U=1/2C*V^2U=1/2(2.07×10^−10 )(7.5^2) = 8.59×10^−6 J
 

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