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: Formula With a Partial Slab:
For a slab of thickness \(t\) and dielectric constant \(K\) in a gap \(d\):
\[ C=\frac{\varepsilon_0A}{d-t+\dfrac tK} \]

Step 2: Substitute:
With \(t=\dfrac{4d}5\): \(d-t=\dfrac d5\), and \(\dfrac tK=\dfrac{4d}{5K}\). So
\[ C=\frac{\varepsilon_0A}{\dfrac d5+\dfrac{4d}{5K}}=\frac{\varepsilon_0A}{d}\cdot\frac{5K}{K+4} \]

Step 3: Ratio:
Without the dielectric \(C_0=\dfrac{\varepsilon_0A}{d}\). So
\[ \frac C{C_0}=\frac{5K}{K+4} \]

Step 4: Check the Options:
For \(K=1\) (no dielectric) the ratio must be 1. Option (C) gives \(5/5=1\). Option (A) gives \(5/5=1\) too, but for large \(K\) the capacitance should approach \(5\,C_0\) because only the \(d/5\) air gap remains. Option (C) tends to 5, while (A) tends to \(1/5\). So (C) is correct.

Final Answer:
The ratio is \(\dfrac{5K}{K+4}\), option (C). \[ \boxed{\text{(C) } \frac{5K}{K+4}} \]