Question

Figure shows a parallel plate air capacitor of plate area of 100 cm² and separation 5 mm. A potential difference of 300 v is established between its plates by a battery.

1. Calculate the capacitance and charge on the capacitor.

2. After disconnecting the battery, the space between the plate Js filled by ebonite (k = 2.6). Then calculate the capacitance and charge on capacitor.

Answer 1

1. \(c=\frac{A\varepsilon_0}{d}\) = \(\frac{100\times10^{-4}\times8.8\times10^{-12}}{5\times10^{-3}}\)

= 17.6 pF

Q = CV= 17.6 × 10^-12 × 300 = 5.2 × 10^-9 C

2. C = \(\frac{AK\varepsilon_0}{d}=\frac{KA\varepsilon_0}{d}\)

= 2.6 × 17.6 × 10^-12 = 45.76pF.