Air capacitor has capacitance C 1 . The space between two plates of capacitor is filled with two dielectric as shown in figure. The new capacitance of the capacitor is C 2 . Then C 2 C 1 is

Question

Air capacitor has capacitance C 1 ​ . The space between two plates of capacitor is filled with two dielectric as shown in figure. The new capacitance of the capacitor is C 2 ​ . Then C 2 ​ C 1 ​ ​ is [ d = distance between two plates, K 1 ​ and K 2 ​ are dielectric constants of first and second dielectrics respectively]

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The initial capacitance of the air capacitor is given by:

$C_{1} = \frac{ϵ _{0} A}{d}$

From the given figure, the two dielectrics are placed in series between the plates. Each dielectric has a thickness of $d /2$ and the same cross-sectional area $A$ .

The capacitance of the first dielectric region is:

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

The capacitance of the second dielectric region is:

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

Since these two capacitors are in series, the equivalent capacitance $C_{2}$ is:

$\frac{1}{C _{2}} = \frac{1}{C _{1}^{'}} + \frac{1}{C _{2}^{'}}$

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

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

Rearranging to find the ratio $\frac{C _{1}}{C _{2}}$ :

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

Substituting the given values $K_{1} = 5$ and $K_{2} = 3$ :

$\frac{C _{1}}{C _{2}} = \frac{5 + 3}{2 \times 5 \times 3} = \frac{8}{30}$

Answer: $\frac{8}{30}$

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