Physics - Current Electricity Question with Solution | TestHub

The capacitance (C) of a parallel plate capacitor with dielectric is given by $\begin{array}{l}C=\frac{{ϵ}_{0}A}{d-t\left(1-\frac{1}{K}\right)}\\ C=100pF=100\times 10^{-12}F,{\in }_0=8.85\times 10^{-12}\\ A=\pi {\mathrm{I}}^2=3.14\times {\left(10^{-2}\right)}^2(\mathrm{r}=\mathrm{lcm})\\ \mathrm{t}=10^{-3}\mathrm{m},\mathrm{d}=\mathrm{t},\mathrm{K}=4\end{array}$ ${\mathrm{C}}_1=\frac{8.85\times 10^{-12}\times 3.14\times 10^{-4}}{10^{-3}-10^{-3}\left(1-\frac{1}{4}\right)}$ $=\frac{27.79\times 10^{-16}}{10^{-3}-0.75\times 10^{-3}}$ $=\frac{27.79\times 10^{-13}}{0.25}$ $=\frac{27.79\times 10^{-13}}{0.25}=111.16\times 10^{-13}\mathrm{F}$ (for 1 set) Required $\mathrm{C}=100\times 10^{-12}\mathrm{F}$ $\Rightarrow \mathrm{n}=\frac{\mathrm{C}}{{\mathrm{C}}_1}=\frac{100\times 10^{-12}}{111.16\times 10^{-13}}\simeq 10$