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Breaking a Capacitor into Combinations

A parallel pl...

Question

A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness $d_{1}$ and dielectric constant $K_{1}$ and the other has thickness $d_{2}$ and dielectric constant $K_{2}$ as shown in figure. This arrangement can be thought as a dielectric slab of thickness d (= $d_{1} + d_{2}$ ) and effective dielectric constant K. The K is

\frac{K_{1} d_{1} + K_{2} d_{2}}{d_{1} + d_{2}}

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\frac{K_{1} d_{1} + K_{2} d_{2}}{K_{1} + K_{2}}

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\frac{K_{1} K_{2} (d_{1} + d_{2})}{{K_{2} d}_{1} + K_{1} d_{2}}

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\frac{{2 K}_{1} K_{2}}{K_{1} + K_{2}}

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Solution

The correct option is C $\frac{K_{1} K_{2} (d_{1} + d_{2})}{{K_{2} d}_{1} + K_{1} d_{2}}$
The capacities of two individual condensers are
$C_{1} = \frac{K_{1} ε_{0} A}{d_{1}}; C_{2} = \frac{K_{2} ε_{0} A}{d_{2}}$
The arrangement is equivalent to two capacitors joined in series

$∴$ Equivalent capacitance, $\frac{1}{C_{e q}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} = \frac{d_{1}}{K_{1} ε_{0} A} + \frac{d_{2}}{K_{2} ε_{0} A}$

$= \frac{1}{ε_{0} A} [\frac{d_{1}}{K_{1}} + \frac{d_{2}}{K_{2}}] = \frac{1}{ε_{0} A} [\frac{{K_{2} d}_{1} + K_{1} d_{2}}{K_{1} + K_{2}}]$

or $C_{e q} = ε_{0} A (\frac{K_{1} + K_{2}}{{K_{2} d}_{1} + K_{1} d_{2}}) . . . . (i)$

Also $C_{e q} = \frac{K ε_{0} A}{d_{1} + d_{2}}$
From (i) and (ii) $ε_{0} A (\frac{K_{1} + K_{2}}{{K_{2} d}_{1} + K_{1} d_{2}}) = ε_{0} A (\frac{K}{d_{1} + d_{2}})$

$∴ K = \frac{K_{1} K_{2} (d_{1} + d_{2})}{K_{1} d_{2} + {K_{2} d}_{1}}$

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