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A parallel plate capacitor having plates...

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

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As is clear from Fig, original capacity,
`C_(1) = (in_(0) S)/(d)`
Figure can be redrawn as shown in

Here, `C = (in_(1) in_(0) S//2)/(d//2) = 2 in_(0) S//d`
`C' = (in_(2) in_(0) S//2)/(d//2) = (4 in_(0) S)/(d)`
As C and C' are in series, their equivalent capacity is `C" = (c' xx C)/(C' + C)`
`:. C" = ((4 in_(0) S)/(d) xx (2 in_(0) S)/(d))/((4 in_(0) S)/(d) + (2 in_(0) S)/(d)) = (4)/(3) in_(0) (S)/(d)`
Now, `C_(3) = (in_(1) in_(0) S//2)/(d) = (in_(0) S)/(d)`
As C" and `C_(3)` are in parallel,
new capacitance, `C_(2) = C" + C_(3)`
`C_(2) = (4)/(3) (in_(0) S)/(d) + (in_(0)S)/(d) = (7)/(3) (in_(0) S)/(d) :. (C_(2))/(C_(1)) = (7)/(3)`
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