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 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 capacitances `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)`