Question

A parallel plate air capacitor has capacity \(C\) and plate area \(A\). Now the space between its plates is filled by two slabs of thickness \(\frac{d}{2}\) each, having dielectric constants \(K_{1}\) and \(K_{2}\). Distance of separation between plates is \(d\). The capacity of the capacitor is (\(\varepsilon _{0}\) = permittivity of free space)

• A. \(\frac{2\varepsilon _{0}d}{A}\left(\frac{K_{1}+K_{2}}{K_{1}K_{2}}\right)\)
• B. \(2C\left(\frac{K_{1}K_{2}}{K_{1}+K_{2}}\right)\)
• C. \(\frac{2\varepsilon _{0}d}{A}(K_{1}+K_{2})\)
• D. \(2C\left(\frac{K_{1}+K_{2}}{K_{1}K_{2}}\right)\)

Hint:

Two slabs of equal thickness stacked between the plates behave as two capacitors in series.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Slabs placed one above the other (each of thickness \(d/2\)) form capacitors in series.

Step 2: Individual Capacitances
\(C_{1}=\dfrac{K_{1}\varepsilon_{0}A}{d/2}=2K_{1}C\) and \(C_{2}=2K_{2}C\), where \(C=\dfrac{\varepsilon_{0}A}{d}\).

Step 3: Series Combination
\(C_{eq}=\dfrac{C_{1}C_{2}}{C_{1}+C_{2}}=\dfrac{4K_{1}K_{2}C^{2}}{2C(K_{1}+K_{2})}=2C\dfrac{K_{1}K_{2}}{K_{1}+K_{2}}\)

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

Final Answer:
The capacitance is \(2C\dfrac{K_{1}K_{2}}{K_{1}+K_{2}}\). \[ \boxed{\text{(B) }2C\left(\dfrac{K_{1}K_{2}}{K_{1}+K_{2}}\right)} \]