Air capacitor has capacitance of 1 μ F . Now the space between two plates of capacitor is filled with two dielectrics as shown in figure. The capacitance of the capacitor is ( d = distance between two

Question

Air capacitor has capacitance of 1 μ F . Now the space between two plates of capacitor is filled with two dielectrics as shown in figure. The capacitance of the capacitor is ( d = distance between two plates, 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:

$C_{0} = \frac{ϵ _{0} A}{d} = 1 μ F$

From the given figure, the space between the plates is divided into two equal halves parallel to the area of the plates. This arrangement is equivalent to two capacitors connected in parallel, each having a plate area of $A /2$ and a plate separation of $d$ .

The capacitance of the upper half filled with dielectric $K_{1}$ is:

$C_{1} = \frac{K _{1} ϵ _{0} ( A /2 )}{d} = \frac{K _{1}}{2} (\frac{ϵ _{0} A}{d}) = \frac{K _{1}}{2} C_{0}$

The capacitance of the lower half filled with dielectric $K_{2}$ is:

$C_{2} = \frac{K _{2} ϵ _{0} ( A /2 )}{d} = \frac{K _{2}}{2} (\frac{ϵ _{0} A}{d}) = \frac{K _{2}}{2} C_{0}$

Since the two capacitors are in parallel, the equivalent capacitance $C$ is the sum of their individual capacitances:

$C = C_{1} + C_{2}$

$C = \frac{K _{1}}{2} C_{0} + \frac{K _{2}}{2} C_{0} = (\frac{K _{1} + K _{2}}{2}) C_{0}$

Substituting the given values $K_{1} = 8$ , $K_{2} = 4$ , and $C_{0} = 1 μ F$ :

$C = (\frac{8 + 4}{2}) \times 1 μ F$

$C = \frac{12}{2} μ F = 6 μ F$

Answer: $6 μ F$

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