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A dielectric-filled parallel-plate capacitor has plate area A-30.0 cm², plate separation d = 8.00 mm and dielectric constant k = 5.00. The capacitor is connected to a battery that cr constant voltage V=7.50 V. Throughout the problem, use co= 8.85x10-12 C²/N m².

A dielectric-filled parallel-plate capacitor has plate area A-30.0 cm², plate separation d = 8.00 mm and dielectric constant k = 5.00. The capacitor is connected to a battery that cr constant voltage V=7.50 V. Throughout the problem, use co= 8.85x10-12 C²/N m².

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I just need part b c and d Part A I got correct which is 4.67*10^-10
**Capacitor Energy Calculation** A dielectric-filled parallel-plate capacitor has the following specifications: - Plate area \( A = 30.0 \, \text{cm}^2 \) - Plate separation \( d = 8.00 \, \text{mm} \) - Dielectric constant \( k = 5.00 \) - Connected to a battery with a constant voltage \( V = 7.50 \, \text{V} \) - Permittivity of free space \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 \) ### Part A **Task:** Find the energy \( U_1 \) of the dielectric-filled capacitor. - Express your answer numerically in joules. **User Input:** \[ U_1 = 11.1 \times 10^{-11} \, \text{J} \] **Feedback:** Incorrect; Try Again; 5 attempts remaining. ### Part B **Task:** When the dielectric plate is half-pulled from the capacitor (still connected to the battery), find the energy \( U_2 \) of the capacitor. - Express your answer numerically in joules. ### Part C The capacitor is disconnected from the battery, and the dielectric plate is slowly removed completely. **Task:** Find the new energy \( U_3 \) of the capacitor. - Express your answer numerically in joules. ### Part D **Task:** Calculate the work \( W \) done by the external agent during the removal of the remaining dielectric from the disconnected capacitor. - Express your answer numerically in joules. Each part includes a section to view available hints, with input fields for answers, and the ability to submit for feedback. The calculations involve understanding capacitance changes with and without a dielectric and energy considerations in capacitors.
Transcribed Image Text:**Capacitor Energy Calculation** A dielectric-filled parallel-plate capacitor has the following specifications: - Plate area \( A = 30.0 \, \text{cm}^2 \) - Plate separation \( d = 8.00 \, \text{mm} \) - Dielectric constant \( k = 5.00 \) - Connected to a battery with a constant voltage \( V = 7.50 \, \text{V} \) - Permittivity of free space \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 \) ### Part A **Task:** Find the energy \( U_1 \) of the dielectric-filled capacitor. - Express your answer numerically in joules. **User Input:** \[ U_1 = 11.1 \times 10^{-11} \, \text{J} \] **Feedback:** Incorrect; Try Again; 5 attempts remaining. ### Part B **Task:** When the dielectric plate is half-pulled from the capacitor (still connected to the battery), find the energy \( U_2 \) of the capacitor. - Express your answer numerically in joules. ### Part C The capacitor is disconnected from the battery, and the dielectric plate is slowly removed completely. **Task:** Find the new energy \( U_3 \) of the capacitor. - Express your answer numerically in joules. ### Part D **Task:** Calculate the work \( W \) done by the external agent during the removal of the remaining dielectric from the disconnected capacitor. - Express your answer numerically in joules. Each part includes a section to view available hints, with input fields for answers, and the ability to submit for feedback. The calculations involve understanding capacitance changes with and without a dielectric and energy considerations in capacitors.
Expert Solution
Step 1: Given

A dielectric-filled parallel plate capacitor with

Plate area

A equals 30.0 space c m squared equals 30.0 space cross times 10 to the power of negative 4 space end exponent m squared

plate separation

d equals 8.00 space m m equals 8.00 cross times 10 to the power of negative 3 space end exponent m

dielectric constant

k equals 5.00

The capacitor is connected to a battery that creates a constant voltage

V equals 7.50 space V

and the permittivity of free space

epsilon subscript o equals 8.85 cross times 10 to the power of negative 12 end exponent space C squared divided by N m squared


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