Energy stored in a capacitor,

$\overline{){\mathbf{U}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{C}}{{\mathbf{V}}}^{{\mathbf{2}}}}$

The capacitance of a parallel plate capacitor,

$\overline{){\mathbf{C}}{\mathbf{=}}\frac{{\mathbf{\epsilon}}_{\mathbf{0}}\mathbf{A}}{\mathbf{d}}}$

**Part A**

Substitute the capacitance in the Energy equation,

${{\mathit{U}}}_{{\mathbf{0}}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{\left(}}\frac{{\mathit{\epsilon}}_{\mathbf{0}}\mathit{A}}{\mathit{d}}{\mathbf{\right)}}{{\mathit{V}}}^{{\mathbf{2}}}$

**Part A. **Find the energy U_{0} stored in the capacitor. Express your answer in terms of A, d, V, and ϵ_{0}.

**Part B. **The capacitor is now disconnected from the battery, and the plates of the capacitor are then slowly pulled apart until the separation reaches 3*d*. Find the new energy *U*_{1} of the capacitor after this process.

Express your answer in terms of *A*, *d*, *V*, and *ϵ*_{0}.

**Part C. **The capacitor is now reconnected to the battery, and the plate separation is restored to *d*. A dielectric plate is slowly moved into the capacitor until the entire space between the plates is filled. Find the energy *U*2 of the dielectric-filled capacitor. The capacitor remains connected to the battery. The dielectric constant is *K*.

Express your answer in terms of *A*, *d*, *V*, *K*, and *ϵ*_{0}.

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