Question

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

• A. \(\text{KC}\)
• B. \(2\text{KC}\)
• C. \(\frac{\text{KC}}{\text{K}+1}\)
• D. \(\frac{2\text{KC}}{\text{K}+1}\)

Hint:

Treat the air gap and the slab as two capacitors in series.

Answer 1

The Correct Option is D

Step 1: Original capacitance
\(C = \frac{\varepsilon_0A}{d}\).

Step 2: Effective separation
With a slab of thickness \(\frac d2\), the effective separation is \(\frac d2 + \frac{d}{2K}\) (air part plus slab thickness divided by \(K\)).

Step 3: New capacitance
\[ C' = \frac{\varepsilon_0A}{\frac d2\left(1+\frac1K\right)} = \frac{2K}{K+1}\cdot\frac{\varepsilon_0A}{d} = \frac{2KC}{K+1} \]
Option (D).

Final Answer:
The capacitance becomes 2KC/(K+1). \[ \boxed{\text{(D)}\ \frac{2KC}{K+1}} \]