A parallel plate capacitor having plates of area S and plate separation d, has capacitance `C_1` in air. When two dielectrics of different relative primitivities (`epsilon_1=2` and `epsilon_2=4`) are introduced between the two plates as shown in the figure, the capacitance becomes `C_2`. The ratio `C_2/C_1` is

Answer

2.33

Explanation

`C_(1) = (epsilon_0S)/(d)`
After insertion of two dielectrics between plates, it becomes a combination of three capacitors as shown in the figure.
`C_(21) to epsilon_(1) = 2, S/2 , d/2 , C_(21) = (2epsilon_0S/2)/(d/2) = (2epsilon_0S)/d = 2C_1`
`C_(23) to epsilon_(1) = 2, S/2 , d/2 , C_(23) = (2epsilon_0S/2)/(d) = (epsilon_0S)/d = C_1`
`C_(22) to epsilon_(2) = 4, S/2 , d/2 , C_(22) = (4epsilon_0S/2)/(d/2) = (4epsilon_0S)/d = 4C_1`
`:. C_(2) = C_(23) + (C_(21) xx C_(22))/(C_(21) + C_(22)) = C_1 + (2C_1 xx 4C_1)/(2C_1 + 4C_1)`
`C_2 = C_1 + 4/3 C_1 = 7/3 C_1 :. (C_2)/(C_1) = 7/3`.