An air-cored capacitor of plate area A and separation d has a capacity C. Two dielectric slabs are inserted between its plates in two different manners as shown. Calculate the capacitance in each case.

Explanation

(i) Let the charges on the plates are Q and -Q.
Electric field in free space is `E_(0)= ( sigma)/(epsilon_(0)) =( Q)/(Aepsilon_(0))`
Electric field in first slab is `E_(1) = ( E_(0))/( K_(1))= ( Q)/(Aepsilon_(0)K_(1))`
Electric field in second slab is `E_(2) = (E_(0))/( K_(2))= ( Q)/( Aepsilon_(0)K_(2)) `
The potential difference between the plates is `V = E_(0)(d-t_(1)-t_(2)) + E_(1)t_(1) + E_(2) t_(2)`
`implies V= E_(0)(d-t_(1)-t_(2)+ ( t_(1))/(K_(1))+(t_(2))/(K_(2)))` ( As, `E_(1) = ( E_(0))/(K_(1)) , E_(2)= ( E_(0))/(K_(2)))`
`:. V= ( Q)/( Aepsilon_(0))(d- t_(1)-t_(2) + (t_(1))/(K_(1))+(t_(2))/(K_(2)))`
`:. C = ( epsilon_(0)A)/( d-t_(1)-t_(2) + ( t_(1))/(K_(1))+ (t_(2))/(K_(2)))`
(ii) The capacitor can be considered as two capacitors of capacitance `C_(1)` and `C_(2)` in parallel

`implies C_(1) = K_(1)(epsilon_(0)A_(1))/(d), C_(2) = ( K_(2)epsilon_(0)A_(2))/(d)`
`implies ` Total capacitance `C = C _(1) +C_(2)`
`implies C = ((K_(1)A_(1)+ K_(2)A_(2))epsilon_(0))/(d)`