A potential dufference of `300V` is applied between the plates of a plane capacitor spaced `1 cm` apart. A plane parallel glass plate with a thickness of `0.5cm` and a plane parallel paraffin plate with a thickness of `0.5 cm` are placed in the space between the capacitor plates find.
(i) Intensity of electric field in each layer.
(ii) The drop of potential in each layer.
(iii) The surface charge density of the charge on capacitor the plates.
Given that `:k_(glass) = 6,,k_("paraffin") = 2`

Answer

(i) `1.5 xx 10^(4)V//m, 4.5 xx 10^(4)V//m`,
(ii) `75V, 225V`,
(iii) `8 xx 10^(-7)C//m^(2)`

Explanation

`C = (in_(0)A)/(d)`

`{:(C_(1) = (6in_(0)A)/(d//2),,rArrC_(1)=12C,,):}`
`{:(C_(2) = (2in_(0)A)/(d//2),,rArrC_(2)=4C,,):}`

`{:(C_(eq) = (4Cxx12C)/(16C),,rArrC_(eq)=3C,,):}`
(ii) Potential will drop in inverse ratio of caapacity
`(V_(1))/(V_(2)) = (4C)/(12C) = (1)/(3)`
`V_(1) = (1)/(4) xx 300 = 75"volt"`,
`V_(2) = (3)/(4) xx 300 "volt" = 225"volt"`
(i) `V_(1) = E_(1)d_(1) rArr E_(1) = 1500 V//m`
`V_(2) = E_(2)d_(2) rArr E_(2) = 4500 V//m`