A parallel plate capacitor with air between the plates has capacitance of $9 p F$ . The separation between its plates is ' $d$ '. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant $k_{1} = 3$ and thickness $\frac{d}{3}$ while the other one has dielectric constant $k_{2} = 6$ and thickness $\frac{2 d}{3}$ . Capacitance of the capacitor is now :

Question

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Answer 1

The air filled capacitance , $C = \frac{A ϵ _{0}}{d} = 9 p F$
According to question there are two capacitor with capacitances $C_{1}$ and $C_{2}$ . and they are in series .
Here, $C_{1} = \frac{A k _{1} ϵ _{0}}{( d / 3 )} = 9 \frac{A ϵ _{0}}{d} = 9 C$ as $k_{1} = 3$
and $C_{2} = \frac{A k _{2} ϵ _{0}}{( 2 d / 3 )} = 9 \frac{A ϵ _{0}}{d} = 9 C$ as $k_{2} = 6$
The equivalent capacitance, $C_{e q} = \frac{C _{1} C _{2}}{C _{1} + C _{2}} = \frac{( 9 C ) ( 9 C )}{9 C + 9 C} = \frac{9}{2} C = (9 / 2) 9 = 40.5 p F$

Answer 2

The air filled capacitance ,

C = \frac{A ϵ _{0}}{d} = 9 p F

According to question there are two capacitor with capacitances

C_{1}

and

C_{2}

. and they are in series .

Here,

C_{1} = \frac{A k _{1} ϵ _{0}}{( d / 3 )} = 9 \frac{A ϵ _{0}}{d} = 9 C

as

k_{1} = 3

and

C_{2} = \frac{A k _{2} ϵ _{0}}{( 2 d / 3 )} = 9 \frac{A ϵ _{0}}{d} = 9 C

as

k_{2} = 6

The equivalent capacitance,

C_{e q} = \frac{C _{1} C _{2}}{C _{1} + C _{2}} = \frac{( 9 C ) ( 9 C )}{9 C + 9 C} = \frac{9}{2} C = (9 / 2) 9 = 40.5 p F