Two parallel palate capacitors `X` and `Y` have the same area of plates and same separation between then. `X` has air between the plates and `Y` contains a dielectric medium of `in_(r) = 4,`
Calculate (i) capacitance of `X` and `Y` if equivalent capacitance fo combination is `4 mu F`. (ii) pot diff between the plates of `X` and `Y`. (iii) What is the ratio of electrostatic energy stored in `X` and `Y` ?

Explanation

(i) Since capacitance (C) of the parallel plate capacitor is given by
`C = (in_(0)A)/(d)`
`:. X = (in_(0)A)/(d)`
and `Y = (in_(0)kA)/(d) = (4 in_(0)A)/(d) " " ( :' k = 4)`
Since X and Y are series combination
`:. (1)/(4) = (1)/(X) + (1)/(Y) rArr (1)/(4) = (1)/(4Y) + (1)/(Y)`
`rArr (1)/(4).(1 + 4)/(4Y) rArr (1)/(4) = (5)/(4Y) :. Y = 5 mu F`
`:. X =4 xx 5 = 20 muF`
(ii) Since, `q = CV " " q = 4 xx 10^(-6) xx 12 = 48 xx 10^(-6) C`
Potential difference on capacitor `X = (q)/(C) = (48 xx 10^(-6))/(20 xx 10^(-6)) = 2.4 V`
and potential difference on capacitor `Y = (48 xx 10^(-6))/(5 xx 10^(-6)) = 9.6 V`
(iii) Since energy stored `= (1)/(2) CV^(2)`
`:.` Ratio of electrostatic energy stored in X and Y `= ((1)/(2) C_(1)V_(1)^(2))/((1)/(2) C_(2) V_(2)^(2))`
`(20 xx (2.4)^(2))/(5 xx (9.6)^(2)) = (4 xx 2.4 xx2.4)/(9.6 xx 9.6) = (4)/(4xx 4) = (1)/(4) = 1 : 4`