Question

A slab of material of dielectric constant \(K\) has the same area as the plates of a parallel plate capacitor but has a thickness \((4/5)d\), where \(d\) is the separation of the plates. The capacitance in the presence and absence of dielectric are \(C\) and \(C_0\) respectively. The ratio \((C/C_0)\) is

• A. \(\frac{K+4}{5K}\)
• B. \(\frac{4K}{K+5}\)
• C. \(\frac{5K}{K+4}\)
• D. \(\frac{K+5}{4K}\)

Hint:

The gap not filled by the slab and the slab itself act like two capacitors in series.

Answer 1

The Correct Option is C

Step 1: Series Combination:
The air gap $d/5$ gives $C_a=\dfrac{5\varepsilon_0A}{d}$. The slab of thickness $4d/5$ gives $C_s=\dfrac{5K\varepsilon_0A}{4d}$.

Step 2: Add in Series:
$\dfrac1C=\dfrac1{C_a}+\dfrac1{C_s}=\dfrac d{5\varepsilon_0A}+\dfrac{4d}{5K\varepsilon_0A}=\dfrac{d}{5\varepsilon_0A}\cdot\dfrac{K+4}{K}$.

Step 3: Result:
$C=\dfrac{5K}{K+4}\cdot\dfrac{\varepsilon_0A}d$. So $C/C_0=\dfrac{5K}{K+4}$. Option (C).

Final Answer:
Option (C). \[ \boxed{\text{(C) } \frac{5K}{K+4}} \]