Question

Find the capacitances of the capacitors shown in figure . The plate area is Aand 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.

Answer 1

The two parts of the capacitor are in series with capacitances C₁ and C₂.
Here,

`C_1 = (K_1∈_0A)/(d/2) and C_2 = (K_2∈_0A)/(d/2)`

⇒ `C_1 = (2K_1∈_0A)/d and C_2 = (2K_2∈_0A)/d`

Because they are in series, the net capacitance is calculated as :

`C = (C_1 xx C_2)/(C_1+C_2)`

= `((2K_1∈_0A)/d xx (2K_2∈_0A)/d)/((2K_1∈_0A)/d xx (2K_2∈_0A)/d)`

= `(2K_1K_2∈_0A)/(d(K_1+K_2)`

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

Now , 

`C_1 = (K_1∈_0A)/((d/3)) = (3K_1∈_0A)/d`

`C_2 = (3K_2∈_0A)/d`

`C_3 = (3K_3∈_0A)/d`

Thus, the net capacitance is calculated as :

`C = (C_1 xx C_2 xx C_2)/(C_1C_2+C_2C_3+C_3C_1)`

 = `((3K_1∈_0A)/d xx (3K_2∈_0A)/d xx (3K_3∈_0A)/d)/((3K_1∈_0A)/d xx (3K_2∈_0A)/d xx (3K_2∈_0A)/d xx (3K_3∈_0A)/d xx (3K_3∈_0A)/d xx (3K_1∈_0A)/d)`

= `(3 ∈_0 K_1K_2K_3)/(d(K_1K_2+K_2K_3+K_3K_1)`

(c)

Here , 

`C_1 = (K_1∈_0A/2)/d = (K_1∈_0A)/(2d)`

`C_2 = (K_2∈_0A)/(2d)`

These two parts are in parallel.

`therefore C = C_1 + C_2`

= `(∈_0A)/(2d)(K_1+K_2)`