Question

wo parallel-plate capacitors have the same dimensions, but the space between the plates is filled with air in capacitor 1 and with plastic in capacitor 2. Each capacitor is connected to an identical battery, such that the potential difference between the plates is the same in both capacitors. Compare the magnitudes of the electric fields between the plates, ????1 and ????2, and the magnitudes of the free charges on the plates, ????1 and ????2.

Answer 1

Answer:

Explanation:

Let the dielectric constant of plastic be K . Then , if the capacity of air capacitor is C , the capacity of plastic capacitor will be KC.

If the emf of battery be V then V will be the potential difference between plates of both of them

electric field E = V/d for air capacitor

For plastic capacitor electric field = E / k

Their ratio

=  E /E / k

Magnitude of free charge in air capacitor

Q₁= V x C= CV

Magnitude of free charge in plastic capacitor

Q₂  = V x KC= KCV

Q₂ / Q₁ = K