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Physics-

General

Easy

Question

A parallel plate capacitor of area A, plate separation d and capacitance $C$ is filled with three different dielectric materials having dielectric constants $K subscript 1 end subscript comma K subscript 2 end subscript blank a n d blank K subscript 3 end subscript$ as shown. If a single dieletric material is to be used to have the same capacitance $C$ is this capacitors, then its dielectric constant $K$ is given by

$\frac{1}{K} = \frac{1}{K_{1}} + \frac{1}{K_{2}} + \frac{1}{2 K_{3}}$
$\frac{1}{K} = \frac{1}{K_{1} {+ K}_{2}} + \frac{1}{2 K_{3}}$
$K = \frac{K_{1} K_{2}}{K_{1} + K_{2}} + 2 K_{3}$
$K = K_{1} + K_{2} + 2 K_{3}$

hint Hint:

A parallel plate capacitor of area A, plate separation d and capacitance $C$ is filled with three different dielectric materials having dielectric constants $K subscript 1 end subscript comma K subscript 2 end subscript blank a n d blank K subscript 3 end subscript$ as shown. If a single dieletric material is to be used to have the same capacitance $C$ is this capacitors, then its dielectric constant $K$ is given by

The correct answer is: $1 over K equals fraction numerator 1 over denominator K subscript 1 plus blank K subscript 2 end fraction plus fraction numerator 1 over denominator 2 K subscript 3 end fraction$

Capacitance of two capacitors each of area $fraction numerator A over denominator 2 end fraction blank comma$ plate separation $d$ but dielectric constants $K subscript 1 end subscript$ and $K subscript 2 end subscript$ respectively joined in parallel
$C subscript 1 end subscript equals fraction numerator K subscript 1 end subscript epsilon subscript 0 end subscript open parentheses fraction numerator A over denominator 2 end fraction close parentheses over denominator d divided by 2 end fraction plus fraction numerator K subscript 2 end subscript epsilon subscript 0 end subscript open parentheses fraction numerator A over denominator 2 end fraction close parentheses over denominator d divided by 2 end fraction equals fraction numerator open parentheses K subscript 1 end subscript plus K subscript 2 end subscript close parentheses epsilon subscript 0 end subscript A over denominator d end fraction$
It is in series with a capacitor of plate area $A comma$ plate separation $d divided by 2 blank$ and dielectric constant $K subscript 3 end subscript blank i e comma blank C subscript 2 end subscript equals fraction numerator K subscript 3 end subscript epsilon subscript 0 end subscript A over denominator d divided by 2 end fraction$ .
If resultant capacitance be taken as $C equals fraction numerator K subscript blank epsilon subscript 0 end subscript A over denominator d end fraction$ ,
Then $fraction numerator 1 over denominator C end fraction equals fraction numerator 1 over denominator C subscript 1 end subscript end fraction plus fraction numerator 1 over denominator C subscript 2 end subscript end fraction$
$therefore fraction numerator d over denominator K epsilon subscript 0 end subscript A end fraction equals fraction numerator d over denominator open parentheses K subscript 1 end subscript plus K subscript 2 end subscript close parentheses epsilon subscript 0 end subscript A end fraction plus fraction numerator d divided by 2 over denominator table row cell K subscript 3 end subscript epsilon subscript 0 end subscript A end cell row cell end cell end table end fraction$
$rightwards double arrow fraction numerator 1 over denominator K end fraction equals fraction numerator 1 over denominator K subscript 1 end subscript plus blank K subscript 2 end subscript end fraction plus fraction numerator 1 over denominator 2 K subscript 3 end subscript end fraction$

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