A parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constant ε 1 and ε 2 , as shown in figures. The distance between the plates is d and area of each plate is A.

Question

A parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constant $ε_{1}$ and $ε_{2}$ , as shown in figures. The distance between the plates is d and area of each plate is A. If capacitance in first configuration and second configuration are $C_{1}$ and $C_{2}$ respectively, then $\frac{C _{1}}{C _{2}}$ is :

TestHub · Accepted Answer

Area of plate is A . then ​ C = d /2 ε 2 ​ ε 0 ​ A ​ = d 2 ε 2 ​ ε 0 ​ A ​ C ′ = d /2 ε 1 ​ ε 0 ​ A ​ = d 2 ε 1 ​ ε 0 ​ A ​ ​ Let C 0 ​ = d ε 0 ​ A ​ ​ C = 2 ε 2 ​ C 0 ​ C ′ = 2 ε 1 ​ C 0 ​ ​ C & C ′ are in series C 1 ​ = C + C ′ CC ′ ​ = 2 C 0 ​ ( ε 2 ​ + ε 1 ​ ) 4 ε 2 ​ ε 1 ​ C 0 2 ​ ​ = ( ε 2 ​ + ε 1 ​ ) 2 ε 2 ​ ε 1 ​ C 0 ​ ​ Here C = 2 d ε 1 ​ ε 0 ​ A ​ = 2 ε 1 ​ C 0 ​ ​ C ′ = 2 ε 2 ​ C 0 ​ ​ C & C ′ are inparallel C 2 ​ = C ′ + C = ( ε 1 ​ + ε 2 ​ ) 2 C 0 ​ ​ Thus C 2 ​ C 1 ​ ​ = ( ε 2 ​ + ε 1 ​ ) 2 ε 2 ​ ε 1 ​ C 0 ​ ​ × ( ε 1 ​ + ε 2 ​ ) C 0 ​ 2 ​ = ( ε 2 ​ + ε 1 ​ ) 2 4 ε 2 ​ ε 1 ​ ​

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