(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 $=\left(5+0.50\right)\mathrm{mm}=5.5\mathrm{mm}$
(i) Effective new capacitance
$=200\mathit{\mu }\mathrm{F}\times \frac{5\mathrm{mm}}{5.5\mathrm{mm}}=\frac{2000}{11}\mathit{\mu }\mathrm{F}$
$\approx 182\mathit{\mu }\mathrm{F}$
(ii) Effective new electric field
$=\frac{100\mathrm{V}}{5.5\times {10}^{-3}\mathrm{m}}=\frac{20000}{1.1}$
$\approx 18182\mathrm{V}∕\mathrm{m}$
(iii) $=\frac{{\mathit{C}}^{\prime }}{\mathit{C}}=\frac{10}{11}$
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 $\mathit{C}=\frac{\mathit{A}{\mathit{\epsilon }}_0}{\mathit{d}}$
${\mathit{C}}_{\mathit{m}}=\frac{\mathit{K}\mathit{A}{\mathit{\epsilon }}_0}{\mathit{d}}$
${\mathit{E}}^{\prime }=\frac{\mathit{V}}{\mathit{d}}$
$\mathit{U}=\frac{1}{2}{\mathit{\epsilon }}_0{\mathit{E}}^2$