Aparallel plate capacitor of area a = 30 cm2 and separation d = 5 mm is charged by a battery of 60-v. if the air between the plates is replaced by a dielectric of κ = 4 with the battery still connected, then what is the ratio of the initial charge on the plates divided by the final charge on the plates?

0.25

Explanation:

A = area of each plate = 30 cm² = 30 x 10⁻⁴ m²

d = separation between the plates = 5 mm = 5 x 10⁻³ m

= Capacitance of capacitor when there is air between the plates

k = dielectric constant = 4

= Capacitance of capacitor when there is dielectric between the plates

Capacitance of capacitor when there is air between the plates is given as

                           eq-1

Capacitance of capacitor when there is dielectric between the plates is given as

                           eq-2

Dividing eq-1 by eq-2

Charge stored in the capacitor when there is air is given as

                            eq-3

Charge stored in the capacitor when there is dielectric is given as

                            eq-4

Dividing eq-3 by eq-4