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Question

# If the two dielectrics of dielectric constant K1 & K2 with thickness t1 & t2 respectively, were to be replaced by a single dielectric completely filling the space between the two plates, then its equivalent dielectric constant will be

A
K1K2(t1+t2)K1t1+K2t2
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B
K1K2(t1+t2)K1t2+K2t1
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C
2K1K2(t1+t2)K1t2+K2t1
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D
K1K2(t1+t2)2(K1t1+K2t1)
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Solution

## The correct option is B K1K2(t1+t2)K1t2+K2t1Suppose the area of the plates is A. The circuit can be redrawn as So, capacitance of both capacitor will be C1=K1ϵ0At1 and C2=K2ϵ0At2 Since both the capacitor are in series, so equivalent capacitance will be Ceq=C1C2C1+C2=K1K2ϵ20A2t1t2ϵ0A(K1t1+K2t2) ⇒Ceq=K1K2ϵ0AK1t2+K2t1 Let Keq be the equivalent dielectric constant which is completely filling the space between the plates. So, capacitance of such capacitor will be C′=Keqϵ0At1+t2 (∵d=t1+t2) Since, the capacitance in both the cases are same, ∴C′=Ceq ⇒Keqϵ0At1+t2=K1K2ϵ0AK1t2+K2t1 ∴Keq=K1K2(t1+t2)K1t2+K2t1 Hence, option (B) is correct. Alternate Solution: Keq=d(t1/K1)+(t2/K2)=K1K2dK2t1+K1t2

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