CUET · PHYSICS · PYQ PAPER 2025
A slab of material of dielectric constant K has the same area as the plates of a parallel-plate capacitor but has a thickness \( \frac{1}{4}d \), where d is the separation between the plates. If the slab is inserted between the plates, the potential difference becomes:
(Given: Potential difference across the capacitor without dielectric is \( V_0 \))
- A \( V = V_0(\frac{3K + 4}{4}) \)
- B \( V = V_0(\frac{3K + 1}{4K}) \)
- C \( V = V_0(\frac{K + 1}{4K}) \)
- D \( V = V_0(\frac{3K + 4}{4K}) \)
Answer & Solution
Correct Answer
(B) \( V = V_0(\frac{3K + 1}{4K}) \)
Step-by-step Solution
Detailed explanation
\( E_0 = \frac{V_0}{d} \) \( V = E_K t + E_{air} (d-t) \) \( V = \frac{E_0}{K} (\frac{d}{4}) + E_0 (\frac{3d}{4}) \) \( V = E_0 d (\frac{1}{4K} + \frac{3}{4}) \) \( V = V_0 (\frac{1 + 3K}{4K}) \)
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