Question

A parallel plate capacitor has plate area \(40\text{ cm}^2\) and plates separation \(2\) mm. The space between the plates is filled with a dielectric medium of a thickness \(1\) mm and dielectric constant \(5\). The capacitance of the system is

• A. \(24ε_0\text{ F}\)
• B. \(\frac{3}{10}ε_0\text{ F}\)
• C. \(\frac{10}{3}ε_0\text{ F}\)
• D. \(10ε_0\text{ F}\)

Hint:

Use $C=\frac{\varepsilon_0A}{d-t+t/K}$.

Answer 1

The Correct Option is C

Step 1: Write the formula
With a dielectric slab of thickness \(t\) and constant \(K\) in a gap \(d\): \(C=\frac{\varepsilon_0A}{d-t+\frac tK}\).

Step 2: Substitute
\(A=40\times10^{-4}\) m\(^2\), \(d=2\times10^{-3}\) m, \(t=10^{-3}\) m, \(K=5\).
Denominator: \(2\times10^{-3}-10^{-3}+\frac{10^{-3}}5=1.2\times10^{-3}\) m.

Step 3: Calculate
\(C=\frac{\varepsilon_0\times4\times10^{-3}}{1.2\times10^{-3}}=\frac{10}{3}\varepsilon_0\) F. Option (C).

Final Answer:
\(C=\frac{10}{3}\varepsilon_0\) F, option (C). \[ \boxed{\text{(C)}} \]