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Transcribed Image Text:**Problem 4.19:** Suppose you have enough linear dielectric material, of dielectric constant \( \epsilon_r \), to *half-fill* a parallel-plate capacitor (Fig. 4.25). By what fraction is the capacitance increased when you distribute the material as in Fig. 4.25(a)? How about Fig. 4.25(b)? For a given potential difference \( V \) between the plates, find \( E \), \( D \), and \( P \), in each region, and the free and bound charge on all surfaces, for both cases.
**Explanation of Figures:**
- **Figure 4.25(a):** The dielectric material is inserted between the plates, with half of the capacitor volume filled from one side of the plates to the other.
- **Figure 4.25(b):** The dielectric material is inserted with half of the capacitor volume filled along the entire area, with the plates parallel but only extending halfway.
In both diagrams, the capacitor plates are shown as parallel plates with the dielectric material filling half the volume, illustrated to help understand different spatial arrangements of dielectric insertion.
For educational purposes, the problem involves understanding how the arrangement of a dielectric within a capacitor affects overall capacitance, electric field (\( E \)), electric displacement (\( D \)), polarization (\( P \)), and charge distribution.
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