Question

A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness $d_1$ and dielectric constant $k_1$ and the other has thickness $d_2$ and dielectric constant $k_2$ as shown in fig. This arrangement can be thought as a dielectric slab of thickness $d=d_1+d_2$ and effective dielectric constant $k$. The $k$ is then :

• A. $\frac{k_1{d}_1+k_2{d}_2}{d_1+d_2}$
• B. $\frac{k_1{d}_1+k_2{d}_2}{k_1+k_2}$
• C. $\frac{k_1{k}_2(d_1+d_2)}{k_1{d}_2+k_2{d}_1}$
• D. $\frac{2k_1{k}_2}{k_1+k_2}$

Answer 1

Answer: (C)

For the first block

Thickness $=d_1$

Dielectric constant $=k_1$

Capacitance $=C_1$

For second block

Thickness $=d_2$

Dielectric constant $=k_2$

Capacitance $=C_2$

Capacitors C₁ and C₂ are connected in parallel. So equivalent capacitance is :
$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}$

$C_{eq}=\frac{C_1{C}_2}{C_1+C_2}........\left(1\right)$

Capacitance in parallel plate capacitor is given as

$C_1=\frac{k_1{\epsilon }_{0}A}{d_1}\&C_2=\frac{k_2{\epsilon }_{0}A}{d_2}$

So, equation (1) becomes

$C_{eq}=\frac{k_1{k}_2{\epsilon }_{0}A}{k_1{d}_2+k_2{d}_1}.............\left(2\right)$

Equivalent capacitance for the combination is

$C=\frac{k{\epsilon }_{0}A}{d_1+d_2}.........\left(2\right)$

From equation (1) and (2)

$\frac{k_1{k}_2{\epsilon }_{0}A}{k_1{d}_2+k_2{d}_1}=\frac{k{\epsilon }_{0}A}{d_1+d_2}$

$k=\frac{k_1{k}_2(d_1+d_2)}{k_1{d}_2+k_2{d}_1}$