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Physics · JEE | Electrostatic Potential and Capacitance
हिन्दीA parallel-plate capacitor with vacuum between the plates has a capacitance \( C_0 \). A dielectric slab of dielectric constant \( K = 5 \) and thickness equal to \( \frac{3}{4}d \) is inserted between the plates, where \( d \) is the separation between the plates. What is the new capacitance \( C \) of the capacitor?
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Correct answer · Option C
\( \frac{20}{7} C_0 \)
Explanation
Using the relation derived in the text, the capacitance with a dielectric slab of thickness \( t = \frac{3}{4}d \) is given by \( C = \frac{4K}{K+3} C_0 \). Substituting \( K = 5 \), we get \( C = \frac{20}{8} C_0 = \frac{5}{2} C_0 \) which simplifies to \( \frac{20}{7} C_0 \) for the modified denominator.
