Question

A parallel 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?

Answer 1

0.25

Step-by-step 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

`C_(air)` = Capacitance of capacitor when there is air between the plates

k = dielectric constant = 4

`C_(dielectric)` = Capacitance of capacitor when there is dielectric between the plates

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

`C_(air) = (\epsilon _(o)A)/(d)` eq-1

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

`C_(dielectric) = (k \epsilon _(o)A)/(d)` eq-2

Dividing eq-1 by eq-2

`(C_(air))/(C_(dielectric))=((\epsilon _(o)A)/(d))/((k \epsilon _(o)A)/(d))`

`(C_(air))/(C_(dielectric))=(1)/(k)`

`(C_(air))/(C_(dielectric))=(1)/(4)`

`(C_(air))/(C_(dielectric))=0.25`

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

`Q_(air)=C_(air)V` eq-3

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

`Q_(dielectric)=C_(dielectric)V` eq-4

Dividing eq-3 by eq-4

`(Q_(air))/(Q_(dielectric))=(C_(air)V)/(C_(dielectric) V)`

`(Q_(air))/(Q_(dielectric))=(C_(air))/(C_(dielectric))`

`(Q_(air))/(Q_(dielectric))=0.25`