Question

A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants $K_1=2,K_2=4,K_3=6$ and $K_4=8$ arranged as shown in the figure. The effective dielectric constant will be

• A. $3.8$
• B. $4.8$
• C. $1.8$
• D. $2.8$

Answer 1

The correct option is B $4.8$


Capacitors having dielectric constants $K_1$ and $K_2$ are in series.
So, the effective capacitance will be

$\frac{1}{C_{e{q}_1}}=\frac{2}{{ϵ}_0{L}^2}(\frac{d}{2K_1}+\frac{d}{2K_2})=\frac{d}{{ϵ}_0{L}^2}(\frac{1}{K_1}+\frac{1}{K_2})$

Since, area $=L\times \frac{L}{2}=\frac{L^2}{2}$

$\Rightarrow \frac{1}{C_{e{q}_1}}=\frac{d}{{ϵ}_0{L}^2}(\frac{1}{2}+\frac{1}{4})=\frac{3d}{4{ϵ}_0{L}^2}$

$\Rightarrow C_{e{q}_1}=\frac{4{ϵ}_0{L}^2}{3d}.........\left(1\right)$

Similarly, for $K_3$ and $K_4$

$C_{e{q}_2}=\frac{24{ϵ}_0{L}^2}{7d}.........\left(2\right)$


Also, Lower and upper halves capacitors are in parallel.
So,

$C_{eq}=C_{e{q}_1}+C_{e{q}_2}=\frac{4{ϵ}_0{L}^2}{3d}+\frac{24{ϵ}_0{l}^2}{7d}$

$\Rightarrow C_{eq}=\frac{100}{21}\frac{{ϵ}_0{L}^2}{d}......\left(1\right)$

Let a dielectric of dielectric constant $K_0$ replace by all four dielectrics. So,

$C_{eq}=\frac{K_{0}A{ϵ}_0}{d}$

Here, Area $=L^2$ and separation $=d$, so we get

$C_{eq}=\frac{K_0{L}^2{ϵ}_0}{d}.......\left(2\right)$

From eq. $\left(1\right)\&\left(2\right)$, we get

$C_{eq}=\frac{100}{21}\frac{{ϵ}_0{L}^2}{d}=\frac{K_0{L}^2{ϵ}_0}{d}$

$\therefore K_0=\frac{100}{21}=4.8$

Hence, option $\left(b\right)$ is correct.