(iii) energy density of the capacitor get affected? Justify your answer in each case. (i) Change in Capacitance (ii) Change in Electric field (iii) Change in Energy density Dielectric slab of thickness $4 \, \mathrm{mm} = \frac{4}{5} \, \mathrm{mm}$ $= 0.8 \, \mathrm{mm}$ Effective (air) plate separation $= (4+0.8) \, \mathrm{mm} = 4.8 \, \mathrm{mm}$ (i) Effective new capacitance $C' = 100 \, \mu \mathrm{F} \times \frac{4 \, \mathrm{mm}}{4.8 \, \mathrm{mm}} = \frac{400}{4.8} \, \mu \mathrm{F}$ Change on capacitor remains same $Q = C \times V = 100 \times 10^{-6} \times 200$ $= 2 \times 10^{-2} \, \mathrm{C}$ New p.d. $V' = \frac{Q}{C'} = \frac{2 \times 10^{-2} \times 4.8 \times 10^{-6}}{400}$ $= 2.4 \times 10^{-6} \, \text{volt}$ (ii) Effective new Electric Field, $E' = \frac{2.4 \times 10^{-10}}{4.8 \times 10^{-3}}$ $= 5 \times 10^{-8} \, \mathrm{Vm}^{-1}$ (iii) $\frac{\text{New energy stored}}{\text{Original energy stored}} = \frac{\frac{1}{2} C' V^{2}}{\frac{1}{2} C V^{2}}$ $\frac{C'}{C} = \frac{400}{4.8 \times 100} = 0.8$ New energy density will be $(0.8)$ of the original density.