Question

The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constant \(K_1\) and \(K_2\) with thickness \(d/2\) and \(d/2\) respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If \(K_1 = 1.25 K_2\), the value of \(K_2\) is :

• A. \( 2.33 \)
• B. \( 1.60 \)
• C. \( 1.33 \)
• D. \( 2.66 \)

Hint:

Treat the capacitor with two dielectrics in series. The equivalent capacitance \(C_{eq}\) is given by \( \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \), where \(C_i = \frac{K_i \epsilon_0 A}{d/2} \). Use the condition \(C_{eq} = 2 C_0\) and the relation \(K_1 = 1.25 K_2\) to solve for \(K_2\).