A parallel plate capacitor with plate area A and plate separation $d = 2 m$ has a capacitance of $4 μ F$ . The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant $K = 3$ (as shown in figure) will be:

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We have, C i ​ = d A ϵ 0 ​ ​ = 4 μ F C f ​ = d − t + k t ​ A ϵ 0 ​ ​ = d − 2 d ​ + 2 × 3 d ​ A ϵ 0 ​ ​ = d ( 1 − 2 1 ​ + 6 1 ​ ) A ∈ 0 ​ ​ = 3 2 ​ 4 μ F ​ = 6 μ F

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