Find the capacitances of the capacitors shown in figure. The plate area is A and the separation between the plates is d. Different dielectric slabs in a particular part of the figure are of the same thickness and the entire gap between the plates is filled with the dielectric slabs.
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