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Question

Three slabs of same dimension having dielectric constant K1,K2 and K3 completely fills the space between the plates of a capacitor as shown in figure. If C is the initial capacitance of the capacitor, the new capacitance is?


A
(K1K2K3K2K3+K1K3+K1K2)3C
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B
(2K1K2K1+K2)C
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C
(K2K3+K1+K3+K1K2K1K2K3)3C
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D
(K1K2K3K1+K2+K3)C
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Solution

The correct option is A (K1K2K3K2K3+K1K3+K1K2)3C
Let the capacitance of dielectric slabs K1,K2 and K3 be C1,C2 and C3 respectively.

C1=K1C0=K1(3ε0Ad)

C2=K2C0=K2(3ε0Ad)

C3=K3C0=K3(3ε0Ad)

From the given figure it is clear that C1,C2 and C3 are in series.

Now capacitance of the new system is,

1Cnew=1C1+1C2+1C3

1Cnew=d3K1ε0A+d3K2ε0A+d3K3ε0A

1Cnew=13K1C+13K2C+13K3C

1Cnew=13C[1K1+1K2+1K3]

Cnew=[K1K2K3K2K3+K1K3+K1K2]3C

Hence, option (a) is the correct answer.

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