# Find the capacita

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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