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

[laterrel]

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=$$\epsilon$$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.

 

[Nate810]

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

 

[Moetasim]

I would first calculate the capacitance of the capacitor in its original state, without the dielectric. Using the equation C = εA/d, where ε is the permittivity of free space (8.85×10−12 C^2/N⋅m^2), A is the plate area (25.0 cm^2), and d is the plate separation (9.00 mm), I would find that the capacitance is 2.78×10−11 F.

Next, I would use the equation C = K*C_0 to find the new capacitance with the dielectric, where K is the dielectric constant (2.00) and C_0 is the original capacitance. This would give a new capacitance of 5.56×10−11 F.

Finally, I would use the equation U=1/2C*V^2 to calculate the energy of the capacitor in the presence of the dielectric. Plugging in the new capacitance and the constant voltage of 7.50 V, I would find that the energy of the capacitor is 2.07×10^−10 J.

I would also note that the area of each portion of the capacitor is half of the original area, since the dielectric fills the space between the plates. This information may be useful for future calculations or experiments involving this capacitor.