Correct 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)`
`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`