# Find the Capacitances of the Capacitors Shown in Figure . the Plate Area is Aand the Separation Between the Plates is D. Different Dielectric Slabs - Physics

Sum

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)

Concept: Capacitors and Capacitance
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 9 Capacitors
Q 56 | Page 169