Question

Two conducting plates, each $3cm\times 6cm$ , and three slabs of dielectric, each $1cm\times 3cm\times 6cm$ and having dielectric constants of $1,2$ , and $3$ , are assembled into a parallel plate capacitor with $d=3cm$ . Determine the two values of capacitance obtained by the two possible methods
of assembling the capacitor. Find $\frac{\left(C_{\text{higher }} - C_{\text{Lower }}\right)}{ϵ_{0}}$ (in $cm$ ). (Approximate the answer to the nearest integer)

Answer 1

$C_{1}=\frac{ϵ_{0} \times 18}{\frac{1}{1} + \frac{1}{2} + \frac{1}{3}}=\frac{ϵ_{0} \times 18}{\frac{6 + 3 + 2}{6}}=ϵ_{0}\times \frac{108}{11}$

$C_{2}=\frac{1 \times ϵ_{0} \times 1 \times 6}{3}+\frac{2 \times ϵ_{0} \times 1 \times 6}{3}+\frac{3 \times ϵ_{0} \times 1 \times 6}{3}$
$=ϵ_{0}\times 12$
$\Rightarrow \frac{C_{2} - C_{1}}{ϵ_{0}}=12\left[1 - \frac{9}{11}\right]=\frac{24}{11}=2.18$