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Question

A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants K1,K2,K3 as shown. If a single dielectric material is to be used to have the same capacitance C in this capacitor then its dielectric constant K is given by
769866_2aaf735010d946f2bfb7115687da12d5.png

A
1K = 1K1+1K2+12K3
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B
1K = 1K1+K2+12K3
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C
1K = K1K2K1+K2+2K3
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D
K = K1K3K1+K3+K2K3K2+K3
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Solution

The correct option is D K = K1K3K1+K3+K2K3K2+K3
Common mistake[Refer Figure 1]:
The common mistake students do in this problem is that they assume C1 and C2 in parallel and their combination in series with C3. Which is completely wrong method. [Refer Fig. 1]

For two capacitors to be taken in parallel, their both ends should be at same potential.
Here, the upper ends of C1 and C2 are at same potential as they are connected to A through a metal plate.
But their lower ends being connected to a dielectric K3 are not at same potential. As Electric field will be different in both dielectrics K1 and K2, since K1 K2.

Step 1: DivideK3region into two equal parts and find capacitance of each parts [Refer Fig. 2]
Hence, To solve this problem, we divide dielectric K3 in two equal parts as shown in figure 2. So,
C1=K1ϵ0(A/2)d/2=K1ϵ0Ad

C2=K2(ϵ0)(A/2)d/2=K2ϵ0Ad

C3=K3ϵ0(A/2)d/2=K3(ϵ0A)d

C4=K3ϵ0(A/2)d/2=K3ϵ0Ad

Now, From figure, Clearly, we can assume C1 and C3 in series and C2 and C4, further their combinations will be in parallel.

Step 2: Equivalent capacitance [Refer Fig. 3]
Combining series combinations in figure 3, we get:

Ceq=C1C3C1+C3+C2C4C2+C4

Putting values of C1,C2,C3 and C4 from Step (1)

Ceq=K1K3K1+K3ϵ0Ad+K2K3K2+K3ϵ0Ad

Step 3: To find Keq, Compare CeqwithKeqϵ0Ad [Refer Fig. 4]
Ceq=Keqϵ0Ad

Keqϵ0Ad=K1K3K1+K3ϵ0Ad+K2K3K2+K3ϵ0Ad

Keq=K1K3K1+K3+K2K3K2+K3

Hence equivalent dielectric constant will be as given in Option D

2111614_769866_ans_63dfd54b65cc49219bfdf779f58d59bd.png

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