Question

A parallel plate capacitor has a plate area of 6x10^-3 m^2 and a plate seperation of 3mm. A pd of 100v is established between its plates by a battery. After disconnecting the battery, the space between the plates is filled by a 3mm thick mica sheet of dielectric constant k=6

Answer 1

A parallel plate capacitor has plate area, A = 6 × 10^-3 m²

seperation between plates , d = 3mm = 3 × 10^-3 m

so, capacitance of parallel plate capacitor , C = $\frac{\epsilon_0A}{d}$

potential difference between plates, V = 100 volts .

now, charge stored in capacitor after disconnecting the battery, Q = CV

= $\frac{\epsilon_0A}{d}100$....(1)

now, after disconnecting the battery, a mica sheet of thickness 3mm is filled between plates.

so, new capacitance, C' = $\frac{\epsilon_0A}{d-t+\frac{t}{K}}$

as here d = t = 3mm

so, $C'=\frac{K\epsilon_0A}{d}$

then, potential difference across capacitor will be , $V=\frac{Q}{C'}$

= $\frac{100}{K}$

putting k = 6

then, potential difference = 100/6 = 16.67 volts.