Question

A parallel plate capacitor has plate area \(50 \mathrm{~cm}^2\) and plate separation 3 mm . The space between the plates is filled with a dielectric medium of thickness 1 mm and dielectric constant 4 . The capacitance becomes ( \(\varepsilon_0=\) permittivity of free space)

• A. \(\frac{18\epsilon _{0}}{7}\)
• B. \(\frac{20\epsilon _{0}}{9}\)
• C. \(\frac{16\epsilon _{0}}{7}\)
• D. \(\frac{14\epsilon _{0}}{5}\)

Hint:

Use C = epsilon_0 A / (d - t + t/K) for a slab of thickness t and dielectric constant K.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
With a dielectric slab of thickness \(t\) and constant \(K\) in a gap \(d\), \(C=\frac{\epsilon_0A}{d-t+\frac{t}{K}}\).

Step 2: Substitute
\(A=50\text{ cm}^2=50\times10^{-4}\text{ m}^2\), \(d=3\text{ mm}\), \(t=1\text{ mm}\), \(K=4\).
Effective gap \(=3-1+\frac{1}{4}=2.25\text{ mm}\).

Step 3: Calculate
\(C=\frac{\epsilon_0\times50\times10^{-4}}{2.25\times10^{-3}}=\frac{20}{9}\epsilon_0\).

Step 4: Compare the options
This is option (B).

Final Answer:
The capacitance becomes \(\frac{20\epsilon_0}{9}\) (in SI units). \[ \boxed{\text{(B) }\dfrac{20\epsilon _{0}}{9}} \]