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.
Solution
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)`
