A capacitor has capacity C when it's parallel plates are separated by air medium of thickness 'd'. A slab of material of dielectric constant K having area equal to that of plates but thickness \(\frac{d}{2}\) is inserted between the plates. Capacitance of the capacitor in the presence of slab will be
Show Hint
- \(\text{KC}\)
- \(2\text{KC}\)
- \(\frac{\text{KC}}{\text{K}+1}\)
- \(\frac{2\text{KC}}{\text{K}+1}\)
The Correct Option is D
Solution and Explanation
Air part: $C_1 = \frac{2\varepsilon_0A}{d} = 2C$. Slab part: $C_2 = \frac{2K\varepsilon_0A}{d} = 2KC$.
Step 2: Combine
$C' = \frac{C_1C_2}{C_1+C_2} = \frac{4KC^2}{2C(1+K)} = \frac{2KC}{K+1}$. Option (D).
Final Answer:
2KC/(K+1). \[ \boxed{\text{(D)}\ \frac{2KC}{K+1}} \]
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