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
Capacity of the parallel plate air capacitor
$=\frac{\varepsilon_0 A}{d}=\frac{8.86 \times 10^{-12} \times 100 \times 10^{-4}}{5 \times 10^{-3}}=1.77 \times 10^{-11} F$
Final capacity of the capacitor with dielectric between the plates is
C' = KC = 2.6 × 1.77 × 10−11, C' = 4.6 × 10−11 F
Initial charge on the capacitor 1.77 × 10−11 × 300 = 5.31 × 10−9 C
Since, the battery has been disconnected, the charge remains the same, therefore the new potential difference is
V' = $\frac{q}{C'}=\frac{5.31 \times 10^{-9}}{4.6 \times 10^{-11}}$ = 115 V
The surface density of charge remains the same in both the cases, i.e.,
$\sigma=\frac{q}{A}=\frac{5.31 \times 10^{-9}}{100 \times 10^{-4}}=5.31 \times 10^{-7} C / m^2$