A parallel-plate air capacitor has a plate area of 100cm³ and separation 5mm. A potential difference of 300V is established between its plates by a battery. After disconnecting a battery, the space between the plates is filled by ebonite (K = 2.6). Find out   initial and final surface-density of charge on the plates.

$ 5.31 \times 10^{-7} C/m^2$

$\text{Capacity of the parallel plate air capacitor} C = \frac{\epsilon_0 A}{d} =\frac{8.86\times 10^{-12}\times 100\times 10^{-4}}{5 \times 10^{-3}}=  1.77\times 10^{-11}F$

                        $\text{Final capacity of the capacitor with dielectric between the plates is}$

                                    $C' = KC = 2.6\times 1.77 \times 10^{-11}F  = 4.6´10^{-11} F$

                        $\text{Initial charge on the capacitor} = 1.77 \times 10^{-11} \times 300 = 5.31 \times 10^{-9}C$

                        $\text{Since, the battery has been disconnected, the charge remains the same, therefore the new potential difference is}$

                                   $ V' = 115V$

                        $\text{The surface density of charge remains the same in both the cases, i.e.,}$

                                    $\sigma =\frac{Q}{A}=5.31 \times 10^{-7}C/m^2$