Air capacitor has capacitance ' C 1 '. The space between two plates of capacitor is filled with two dielectrics as shown in figure. The new capacitance of the capacitor is ' C 2 &#0

Question

Air capacitor has capacitance ' C 1 ​ '. The space between two plates of capacitor is filled with two dielectrics as shown in figure. The new capacitance of the capacitor is ' C 2 ​ '. The ratio C 2 ​ C 1 ​ ​ is ( d = distance between two plates of capacitor, K 1 ​ and K 2 ​ are dielectric constants of two dielectrics respectively)

TestHub · Accepted Answer

The initial capacitance of the air capacitor is given by:

$C_{1} = \frac{ϵ _{0} A}{d}$

where $A$ is the area of the plates and $d$ is the distance between them.

From the given figure, the space between the plates is filled with two dielectrics of dielectric constants $K_{1}$ and $K_{2}$ , each occupying half the distance between the plates (i.e., $d /2$ ). This arrangement is equivalent to two capacitors connected in series.

The capacitance of the first part 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 part is:

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

The equivalent capacitance $C_{2}$ of these two capacitors in series is:

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

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

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

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

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

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

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

Physics - Capacitor Question with Solution | TestHub

Options:

Doubts & Discussion