A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants $k_{1}, k_{2}$ and $k_{3}$ 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
[Take k₁=2 , k₂=4 , k₃=6 ]

Question

A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants $k_{1}, k_{2}$ and $k_{3}$ 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

[Take k₁=2 , k₂=4 , k₃=6 ]

Infinity Learn · Accepted Answer

$$C1=k1$$∈$$0A2d2=k1$$∈$$0Ad$$ $$C2=k2$$∈$$0A2d2=k2$$∈$$0Ad$$ and $$C3=k3$$∈$$0A2d2=k3$$∈$$0Ad$$  Ceff $$=$$ $$C1C3C1+C3$$ $$+$$ $$C2C3C2+C3$$ ⇒Ceff $$=$$ ∈$$0Adk1k3k1+k3+k2k3k2+k3$$ Hence, keff $$=$$ $$k1k3k1+k3+k2k3k2+k3=128+2410=3.9$$

Insertion of dielectric in capacitor

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Insertion of dielectric in capacitor

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C1=k1∈0A2d2=k1∈0Ad C2=k2∈0A2d2=k2∈0Ad and C3=k3∈0A2d2=k3∈0Ad Ceff = C1C3C1+C3 + C2C3C2+C3 ⇒Ceff = ∈0Adk1k3k1+k3+k2k3k2+k3 Hence, keff = k1k3k1+k3+k2k3k2+k3=128+2410=3.9

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