A parallel plate capacitor of area , plate separation and capacitance is filled with three different dielectric materials having dielectric constants , and , as shown. If a single dielectric material is to be used to have the same capacitance in this

Question

A parallel plate capacitor of area $A$ , plate separation $d$ and capacitance $C$ is filled with three different dielectric materials having dielectric constants $k_{1}$ , $k_{2}$ and $k_{3}$ , as shown. If a single dielectric material is to be used to have the same capacitance $C$ in this capacitor, then its dielectric constant $k$ is given by [2000]

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(2)

Let $d$ , $C_{2}$ and $C_{3}$ are the capacitances of capacitors with dielectric constants $K_{1}$ , $K_{2}$ and $K_{3}$ respectively.

$∴$ $C_{1} = K_{1} (\frac{A}{2}) \frac{ε_{0} \times 2}{d} = \frac{A ε_{0} K_{1}}{d}$

$∴$ $C_{2} = K_{2} (\frac{A}{2}) \frac{ε_{0} \times 2}{d} = \frac{A ε_{0} K_{2}}{d}$

$∴$ $C_{3} = K_{3} (A) \frac{ε_{0} \times 2}{d} = \frac{2 A ε_{0} K_{3}}{d}$

$C_{1}$ and $C_{2}$ are in parallel.

$∴$ $C_{eq} = \frac{A ε_{0}}{d} (K_{1} + K_{2})$

Again, this $C_{eq}$ and $C_{3}$ are in series.

$∴$ $\frac{1}{C} = \frac{d}{A ε_{0} (K_{1} + K_{2})} + \frac{d}{2 A ε_{0} K_{3}}$

But $C = \frac{K A ε_{0}}{d}$ for a single equivalent capacitor.

$∴$ $\frac{d}{K A ε_{0}} = \frac{d}{A ε_{0} (K_{1} + K_{2})} + \frac{d}{2 A ε_{0} K_{3}}$

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

Solution

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

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

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

4 $K = K_{1} + K_{2} + 2 K_{3}$

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Solution

1 1K=1K1+1K2+12K3

2 1K=1K1+K2+12K3

3 K=K1K2K1+K2+2K3

4 K=K1+K2+2K3

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

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

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

4 $K = K_{1} + K_{2} + 2 K_{3}$