Answer :

a)The capacitors are as shown in the fig



Here, the two parts of the capacitor


are in series


Capacitances C1 and C2 with dielectric constants as K1 and K2



Where,


A= Plate Area


d= separation between the plates,


0 = Permittivity of free space = 8.854 × 10-12 m-3 kg-1 s4 A2


k = dielectric strengthof the material


Here,


and


Since, the distance between the plates is divided into two parts,


hence, separation between the plates becomes =


and



Because they are in series, the equivalent capacitance is


calculated as:



Substituting the values,




Here, the capacitor has three parts. These can be taken in series.


b)



Now, in this case, there are three capacitors connected as shown in fig.


These capacitors are connected in series.


capacitance c is given by –



Where,


A= Plate Area


d= separation between the plates,


0 = Permittivity of free space = 8.854 × 10-12 m-3 kg-1 s4 A2


k = dielectric strengthof the material


Capacitors are as follows –


Since, the entire distance is separated into three parts,




Similarly, the other two capacitors




These three capacitors are connected in series


Thus, the net capacitance is calculated as-






c) Here, the capacitors are connected as shown in fig.



We know, capacitance c is given by-



Where,


A= Plate Area


d= separation between the plates,


0 = Permittivity of free space = 8.854 × 10-12 m-3 kg-1 s4 A2


k = dielectric strengthof the material


Capacitors C1 andC2 is given by-





These two capacitors are connected in parallel, net capacitance


becomes




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