A parallel plate air capacitor has capacitance C p . It is equally filled with parallel layers of materials of dielectric constants K 1 and K 2 . Now its capacity becomes C K . The ratio C P

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A parallel plate air capacitor has capacitance C p ​ . It is equally filled with parallel layers of materials of dielectric constants K 1 ​ and K 2 ​ . Now its capacity becomes C K ​ . The ratio C P ​ to C K ​ is

TestHub · Accepted Answer

The initial capacitance of the parallel plate air capacitor is given by:

$C_{P} = \frac{ϵ _{0} A}{d}$

When the capacitor is equally filled with parallel layers of dielectric materials, the layers are parallel to the plates. This arrangement acts as two capacitors connected in series, each having the same area $A$ but half the thickness $\frac{d}{2}$ .

The capacitance of the first layer is:

$C_{1} = \frac{K _{1} ϵ _{0} A}{\frac{d}{2}} = \frac{2 K _{1} ϵ _{0} A}{d} = 2 K_{1} C_{P}$

The capacitance of the second layer is:

$C_{2} = \frac{K _{2} ϵ _{0} A}{\frac{d}{2}} = \frac{2 K _{2} ϵ _{0} A}{d} = 2 K_{2} C_{P}$

The equivalent capacitance $C_{K}$ for the series combination is given by:

$\frac{1}{C _{K}} = \frac{1}{C _{1}} + \frac{1}{C _{2}}$

$\frac{1}{C _{K}} = \frac{1}{2 K _{1} C _{P}} + \frac{1}{2 K _{2} C _{P}}$

$\frac{1}{C _{K}} = \frac{K _{1} + K _{2}}{2 K _{1} K _{2} C _{P}}$

Rearranging to find the ratio $\frac{C _{P}}{C _{K}}$ :

$\frac{C _{P}}{C _{K}} = \frac{K _{1} + K _{2}}{2 K _{1} K _{2}}$

Answer: $\frac{K _{1} + K _{2}}{2 K _{1} K _{2}}$

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