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.
Answer & explanation
Correct answer: option 3
$\text{Capacity of the parallel plate air capacitor} C = \frac{\epsilon_0 A}{d} =\frac{8.86\times 10^{-12}\times 100\times 10^{-4}}{5 \times 10^{-3}}= 1.77\times 10^{-11}F$
$\text{Final capacity of the capacitor with dielectric between the plates is}$
$C' = KC = 2.6\times 1.77 \times 10^{-11}F = 4.6´10^{-11} F$
$\text{Initial charge on the capacitor} = 1.77 \times 10^{-11} \times 300 = 5.31 \times 10^{-9}C$
$\text{Since, the battery has been disconnected, the charge remains the same, therefore the new potential difference is}$
$ V' = 115V$
$\text{The surface density of charge remains the same in both the cases, i.e.,}$
$\sigma =\frac{Q}{A}=5.31 \times 10^{-7}C/m^2$