Question

The space between the plates of a parallel plate capacitor of capacitance $9\mu \text{F}$ is filled with three dielectric slabs of identical size as shown in the figure. The new capacitance is

• A. $15\mu \text{F}$
• B. $20\mu \text{F}$
• C. $25\mu \text{F}$
• D. $27\mu \text{F}$

Answer 1

The correct option is D $27\mu \text{F}$

Considering each one third of the assembly as a separate capacitor. The three positive plates are connected together at point $\text{A}$ and the three negative plates are connected together at point $\text{B}$. Thus, the three capacitors are joined in parallel.

We know that for parallel combination of capacitors, equivalent capacitance is given by

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

Since, the plate area is one third of the original for each part

$C_{eq}=\frac{{ϵ}_0}{d}(\frac{A}{3}{K}_1+\frac{A}{3}{K}_2+\frac{A}{3}{K}_3)$

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

Given: $C=\frac{A{ϵ}_0}{d}=9\mu \text{F}$

$\therefore C_{eq}=\frac{C}{3}(K_1+K_2+K_3)$

$\Rightarrow C_{eq}=\frac{9}{3}(2+3+4)=27\mu \text{F}$

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

Why this question? Trick - For combination of capacitors, first find the types of combination (series or parallel) then apply the relevant formula.