Question

A parallel plate capacitor of plate separation \(2\) mm and plate area \(40\) \(\text{cm}^{2}\) is filled with dielectric whose area is same as plates of capacitor. Thickness is \(1\) mm and dielectric constant is \(5\). New capacitance is (\(\epsilon _{0}\) = permittivity of free space)

• A. \(\frac{3\epsilon _{0}}{10}F\)
• B. \(\frac{10\epsilon _{0}}{3}F\)
• C. \(\frac{20\epsilon _{0}}{3}F\)
• D. \(\frac{3\epsilon _{0}}{20}F\)

Hint:

With a slab of thickness t and dielectric constant K, use C equal to epsilon naught A over d minus t plus t over K.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
With a dielectric slab of thickness \(t\) between plates separated by \(d\):
\(C=\dfrac{\epsilon_0A}{d-t+\dfrac{t}{K}}\).

Step 2: Convert units
\(A=40\ \text{cm}^2=40\times10^{-4}\ \text{m}^2\), \(d=2\times10^{-3}\) m, \(t=1\times10^{-3}\) m, \(K=5\).

Step 3: Effective separation
\(d-t+\dfrac{t}{K}=10^{-3}+0.2\times10^{-3}=1.2\times10^{-3}\) m.

Step 4: Calculation
\(C=\dfrac{\epsilon_0\times40\times10^{-4}}{1.2\times10^{-3}}=\dfrac{10\epsilon_0}{3}\) F, option (B).

Final Answer:
The new capacitance is 10 epsilon naught by 3 farad. \[ \boxed{\text{(B) }\dfrac{10\epsilon_{0}}{3}\,F} \]