The plates of a parallel plate capacitor are separated by $d$. Two slabs of different dielectric con

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 time larger than when there is nothing between the plates. If $K_1=1.25 K_2$, the value of $K_1$ is

Professor PC Thomas Classes · Answer

(c) : Capacitance without dielectric 
$ C_0=\varepsilon_0 \frac{A}{d} $ 
Capacitance with first dielectric slab, 
$ C_1=\frac{K_1 \varepsilon_0 A}{\frac{3}{8} d}=\frac{8 K_1 \varepsilon_0 A}{3 d} $ 
Capacitance with second dielectric slab, 
$ C_2=\frac{K_2 \varepsilon_0 A}{\frac{d}{2}}=\frac{2 K_2 \varepsilon_0 A}{d} $ 
Capacitance with air, 
$ C_3=\frac{\varepsilon_0 A}{\left(d-\frac{3 d}{8}-\frac{d}{2}ight)}=\frac{\varepsilon_0 A 	imes 8}{d} $
 Equivalent capacitance, $\frac{1}{C_{e q}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$ 
$ \begin{aligned} & 	ext { or } \frac{1}{C_{e q}}=\frac{3 d}{8 K_1 \varepsilon_0 A}+\frac{d}{2 K_2 \varepsilon_0 A}+\frac{d}{\varepsilon_0 A 	imes 8} \ & 	ext { or } \frac{1}{C_{e q}}=\frac{d}{\varepsilon_0 A}\left[\frac{3}{8 K_1}+\frac{1}{2 K_2}+\frac{1}{8}ight] \end{aligned} $
 Given $C_{	ext {eq }}=2 C_0$ 
$ 	ext { or } \quad \frac{C_0}{C_{\mathrm{eq}}}=\frac{1}{2} $
 $\begin{aligned} & 	herefore \frac{\varepsilon_0 A}{d} 	imes \frac{d}{\varepsilon_0 A}\left[\frac{3}{8 K_1}+\frac{1}{2 K_2}+\frac{1}{8}ight]=\frac{1}{2} \ & \Rightarrow \frac{3}{8 	imes 1.25 K_2}+\frac{1}{2 K_2}=\frac{1}{2}-\frac{1}{8} \ & \Rightarrow \frac{3}{10 K_2}+\frac{1}{2 K_2}=\frac{3}{8} \ & \Rightarrow \frac{3+5}{10 K_2}=\frac{3}{8} \ & \Rightarrow 64=30 K_2 \ & \Rightarrow K_2=\frac{64}{30}=2.13=\frac{K_1}{1.25} \Rightarrow K_1=2.66\end{aligned}$