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

AI Thread Summary
The discussion focuses on calculating the energy stored in a dielectric-filled parallel-plate capacitor with a specified plate area, separation, and voltage. The dielectric constant is given as 2.00, and the relevant equations for capacitance and energy are provided. The capacitance is calculated using the formula C=K*C_0, resulting in a value of 2.07×10^−10 F. Subsequently, the energy stored in the capacitor is determined to be 8.59×10^−6 J. The calculations demonstrate the impact of the dielectric on capacitance and energy storage.
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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=\epsilonA/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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