The plates of a parallel plate capacitor are separated by . Two slabs of different dielectric constants and with thickness and , respectively, are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothi

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Solution

Q.
The plates of a parallel plate capacitor are separated by $d$ . Two slabs of different dielectric constants $K_{1}$ and $K_{2}$ with thickness $\frac{3}{8} d$ and $\frac{d}{2}$ , respectively, are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates.

If $K_{1} = 1.25 K_{2}$ , the value of $K_{1}$ is [2025]

1 1.60

2 1.33

3 2.66

4 2.33

Ans.
(3)

Capacitance without dielectric

$C_{0} = ε_{0} \frac{A}{d}$

Capacitance with first dielectric slab,

$C_{1} = \frac{K_{1} ε_{0} A}{\frac{3}{8} d} = \frac{8 K_{1} ε_{0} A}{3 d}$

Capacitance with second dielectric slab,

$C_{2} = \frac{K_{2} ε_{0} A}{\frac{d}{2}} = \frac{2 K_{2} ε_{0} A}{d}$

Capacitance with air,

$C_{3} = \frac{ε_{0} A}{(d - \frac{3 d}{8} - \frac{d}{2})} = \frac{ε_{0} A \times 8}{d}$

Equivalent Capacitance, $\frac{1}{C_{e q}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$

or    $\frac{1}{C_{e q}} = \frac{3 d}{8 K_{1} ε_{0} A} + \frac{d}{2 K_{2} ε_{0} A} + \frac{d}{ε_{0} A \times 8}$

or    $\frac{1}{C_{e q}} = \frac{d}{ε_{0} A} [\frac{3}{8 K_{1}} + \frac{1}{2 K_{2}} + \frac{1}{8}]$ ...(ii)

Given $C_{e q} = 2 C_{0}$

or $\frac{C_{0}}{C_{e q}} = \frac{1}{2}$

$∴$    $\frac{ε_{0} A}{d} \times \frac{d}{ε_{0} A} [\frac{3}{8 K_{1}} + \frac{1}{2 K_{2}} + \frac{1}{8}] = \frac{1}{2}$

$\Rightarrow \frac{3}{8 \times 1.25 K_{2}} + \frac{1}{2 K_{2}} = \frac{1}{2} - \frac{1}{8}$

$\Rightarrow \frac{3}{10 K_{2}} + \frac{1}{2 K_{2}} = \frac{3}{8}$

$\Rightarrow \frac{3 + 5}{10 K_{2}} = \frac{3}{8}$

$\Rightarrow 64 = 30 K_{2}$

$\Rightarrow K_{2} = \frac{64}{30} = 2.13 \Rightarrow \frac{K_{1}}{1.25} \Rightarrow K_{1} = 2.66$

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