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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.
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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)`
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