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 \(\frac{3}{8} d\) and \(\frac{d}{2}\), 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 \(K_{1}\) is:

(1) 2.66

(2) 2.33

(3) 1.60

(4) 1.33

Answer 1

Correct option is : (1) 2.66

\(=\text{Using}\ C_{e q}=\frac{\varepsilon_{0} A}{\frac{t_{1}}{\kappa_{1}}+\frac{t_{2}}{K_{2}}+\frac{t_{3}}{\kappa_{3}}}\)

\(\text{here}\ C_{0}=\frac{\varepsilon_{0} A}{d}, t_{1}=\frac{3 d}{8}, t_{2}=\frac{d}{2}, t_{3}=\frac{d}{8}\)

\( K_{1}=K_{1}, \quad K_{2}=\frac{K_{1}}{1.25} \ \text{and} \ K_{3}=1\)

Given \(C_{\text {eq }}=2 C_{0}\)

\( \begin{aligned} \Rightarrow 2 C_{0}=\frac{\varepsilon_{0} A}{\frac{3 d}{8 K_{1}}+\frac{d \times 1.25}{2 K_{1}}+\frac{d}{8}} \end{aligned}\)

\( \begin{aligned} & \Rightarrow \frac{2 \varepsilon_{0} A}{d}=\frac{\varepsilon_{0} A}{\frac{3 d}{8 K_{1}}+\frac{d}{2 K_{1}} \times \frac{5}{4}+\frac{d}{8}} \end{aligned} \)

\( \begin{aligned} \Rightarrow 2=\frac{1}{\frac{3}{8 K_{1}}+\frac{5}{8 K_{1}}+\frac{1}{8}} \Rightarrow K_{1}=\frac{8}{3}=2.66 \end{aligned} \)