First slide

Insertion of dielectric in capacitor

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 ]

Moderate

Solution

$C_{1} = \frac{k_{1} \in_{0} \frac{A}{2}}{(\frac{d}{2})} = \frac{k_{1} \in_{0} A}{d} C_{2} = \frac{k_{2} \in_{0} \frac{A}{2}}{(\frac{d}{2})} = \frac{k_{2} \in_{0} A}{d} a n d C_{3} = \frac{k_{3} \in_{0} \frac{A}{2}}{(\frac{d}{2})} = \frac{k_{3} \in_{0} A}{d} C_{e f f} = \frac{C_{1} C_{3}}{C_{1} + C_{3}} + \frac{C_{2} C_{3}}{C_{2} + C_{3}} \Rightarrow C_{e f f} = \frac{\in_{0} A}{d} (\frac{k_{1} k_{3}}{k_{1} + k_{3}} + \frac{k_{2} k_{3}}{k_{2} + k_{3}}) H e n c e, k_{e f f} = \frac{k_{1} k_{3}}{k_{1} + k_{3}} + \frac{k_{2} k_{3}}{k_{2} + k_{3}} = \frac{12}{8} + \frac{24}{10} = 3.9$

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