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

Answer: 2

$C_1=\frac{{ϵ}_0\times 18}{\frac{1}{1}+\frac{1}{2}+\frac{1}{3}}=\frac{{ϵ}_0\times 18}{\frac{6+3+2}{6}}={ϵ}_0\times \frac{108}{11}$

$C_2=\frac{1\times {ϵ}_0\times 1\times 6}{3}+\frac{2\times {ϵ}_0\times 1\times 6}{3}+\frac{3\times {ϵ}_0\times 1\times 6}{3}$
$={ϵ}_0\times 12$
$\Rightarrow \frac{C_2-C_1}{{ϵ}_0}=12\left[1-\frac{9}{11}\right]=\frac{24}{11}=2.18$