(i) Change in capacitance (ii) Change in electric field (iii) Change in electric density Dielectric slab of thickness $5\;\mathrm{mm}$ is equivalent to an air capacitor of thickness $=\frac{5}{10}\;\mathrm{mm}$ Effective separation between the plates with air in between is $=(5+0.50)\;\mathrm{mm}=5.5\;\mathrm{mm}$ (i) Effective new capacitance $\;=200\;\mu\mathrm{F}\times\frac{5\;\mathrm{mm}}{5.5\;\mathrm{mm}}=\frac{2000}{11}\;\mu\mathrm{F}$ $\;\;\approx 182\;\mu\mathrm{F}$ (ii) Effective new electric field $\;=\frac{100\;\mathrm{V}}{5.5\times10^{-3}\;\mathrm{m}}=\frac{20000}{1.1}$ $\;\;\approx 18182\;\mathrm{V}/\mathrm{m}$ (iii) $\frac{\text{ New energy stored }}{\text{ Original energy stored }}=\frac{\frac{1}{2}C' V^2}{\frac{1}{2}C V^2}$ $=\frac{C'}{C}=\frac{10}{11}$ New Energy density will be $\left(\frac{10}{11}\right)^2$ of the original energy density $=\frac{100}{121}$ of the original energy density. [Note: If the student writes $C=\frac{A\varepsilon_0}{d}$$C_m=\frac{K A\varepsilon_0}{d}$$E'=\frac{V}{d}$$U=\frac{1}{2}\varepsilon_0 E^2$