A parallel plate condenser with plate area A and separation d is filled with two dielectric materials as shown in the given below. The dielectric constants are K1 and K2 respectively. The capacitance will be:-

Question

A parallel plate condenser with plate area A and separation d is filled with two dielectric materials as shown in the given below. The dielectric constants are K1 and K2 respectively. The capacitance will be:- <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-mjqfaihqy0vhyz8l.jpg" /> (A) ((ϵ )0 A/ d)(K1 + K2) (B) ((ϵ )0 A/ d)((K1 + K2/K1 K2)) (C) (2 (ϵ )0 A/ d)((K1 K2/ K1 + K2)) (D) (2 (ϵ )0 A/d)((K1 + K2/ K1 K2))

Tardigrade Admin · Accepted Answer

The combination is equivalent to two capacitors in series, each with plate area

A

and separation

d /2

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

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

Further,

\frac{1}{C} = \frac{1}{C _{1}} + \frac{1}{C _{2}} = \frac{d}{2 ( ϵ ) _{0} A} (\frac{1}{K _{1}} + \frac{1}{K _{2}})

Q. A parallel plate condenser with plate area $A$ and separation $d$ is filled with two dielectric materials as shown in the given below. The dielectric constants are $K_{1}$ and $K_{2}$ respectively. The capacitance will be:-

$\frac{( ϵ ) _{0} A}{d} (K_{1} + K_{2})$

$\frac{( ϵ ) _{0} A}{d} (\frac{K _{1} + K _{2}}{K _{1} K _{2}})$

$\frac{2 ( ϵ ) _{0} A}{d} (\frac{K _{1} K _{2}}{K _{1} + K _{2}})$

$\frac{2 ( ϵ ) _{0} A}{d} (\frac{K _{1} + K _{2}}{K _{1} K _{2}})$

Q. A parallel plate condenser with plate area A and separation d is filled with two dielectric materials as shown in the given below. The dielectric constants are K1​ and K2​ respectively. The capacitance will be:-

d(ϵ)0​A​(K1​+K2​)

d(ϵ)0​A​(K1​K2​K1​+K2​​)

d2(ϵ)0​A​(K1​+K2​K1​K2​​)

d2(ϵ)0​A​(K1​K2​K1​+K2​​)

The combination is equivalent to two capacitors in series, each with plate area A and separation d/2 C1​=(d/2)K1​(ϵ)0​A​=d2K1​(ϵ)0​A​ C2​=(d/2)K2​(ϵ)0​A​=d2K2​(ϵ)0​A​ Further, C1​=C1​1​+C2​1​=2(ϵ)0​Ad​(K1​1​+K2​1​)

Q. A parallel plate condenser with plate area $A$ and separation $d$ is filled with two dielectric materials as shown in the given below. The dielectric constants are $K_{1}$ and $K_{2}$ respectively. The capacitance will be:-

$\frac{( ϵ ) _{0} A}{d} (K_{1} + K_{2})$

$\frac{( ϵ ) _{0} A}{d} (\frac{K _{1} + K _{2}}{K _{1} K _{2}})$

$\frac{2 ( ϵ ) _{0} A}{d} (\frac{K _{1} K _{2}}{K _{1} + K _{2}})$

$\frac{2 ( ϵ ) _{0} A}{d} (\frac{K _{1} + K _{2}}{K _{1} K _{2}})$