Question

A parallel plate capacitor has area $6{\text{cm}}^2$ and separation $3\text{mm}$. The gap between the plates is filled with three dielectric materials of equal thickness as shown. The equivalent dielectric constant of a material which would give the same capacitance when fully inserted in the capacitor, would be

• A. $4$
• B. $36$
• C. $12$
• D. $14$

Answer 1

The correct option is C $12$


From figure, we can see that the capacitors are in parallel.

So, the equivalent capacitor is given by

$C_{eq}=\frac{{ϵ}_0}{d}(A_1{K}_1+A_2{K}_2+A_3{K}_3)$

Here the plate are is one third of the original for each part.
So, $A_1=A_2=A_3=A/3$

$\Rightarrow C_{eq}=\frac{{ϵ}_{0}A}{3d}(K_1+K_2+K_3)$

Here, $K_1=10;K_2=12;K_3=14$

$\Rightarrow C_{eq}=\frac{36{ϵ}_{0}A}{3d}=\frac{12{ϵ}_{0}A}{d}.......\left(1\right)$

Suppose, all three dielectric slab is replaced by a material of dielectric constant $K_0$, which is fully inserted in the capacitor, then,

$C=\frac{K_0{ϵ}_{0}A}{d}..........\left(2\right)$

As the capacitance is both the cases are equal,
so,

$C_{eq}=C$

$\Rightarrow \frac{12{ϵ}_{0}A}{d}=\frac{K_0{ϵ}_{0}A}{d}$

$\Rightarrow K_0=12$

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

$$\begin{array}{c}\text{Why this question ?}\\ \text{Caution- some data are given in question}\\ \text{to confuse you. (Like plate area and}\\ \text{plate separation).}\end{array}$$