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Kirchhoff's Voltage Law

If the two di...

Question

If the two dielectrics of dielectric constant $K_{1}$ & $K_{2}$ with thickness $t_{1}$ & $t_{2}$ respectively, were to be replaced by a single dielectric completely filling the space between the two plates, then its equivalent dielectric constant will be

\frac{K_{1} K_{2} (t_{1} + t_{2})}{K_{1} t_{1} + K_{2} t_{2}}

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\frac{K_{1} K_{2} (t_{1} + t_{2})}{K_{1} t_{2} + K_{2} t_{1}}

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\frac{2 K_{1} K_{2} (t_{1} + t_{2})}{K_{1} t_{2} + K_{2} t_{1}}

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\frac{K_{1} K_{2} (t_{1} + t_{2})}{2 (K_{1} t_{1} + K_{2} t_{1})}

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Solution

The correct option is B $\frac{K_{1} K_{2} (t_{1} + t_{2})}{K_{1} t_{2} + K_{2} t_{1}}$
Suppose the area of the plates is $A$ . The circuit can be redrawn as

So, capacitance of both capacitor will be

$C_{1} = \frac{K_{1} ϵ_{0} A}{t_{1}}$ and $C_{2} = \frac{K_{2} ϵ_{0} A}{t_{2}}$

Since both the capacitor are in series, so equivalent capacitance will be

$C_{e q} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = \frac{K_{1} K_{2} ϵ_{0}^{2} A^{2}}{t_{1} t_{2} ϵ_{0} A (\frac{K_{1}}{t_{1}} + \frac{K_{2}}{t_{2}})}$

$\Rightarrow C_{e q} = \frac{K_{1} K_{2} ϵ_{0} A}{K_{1} t_{2} + K_{2} t_{1}}$

Let $K_{e q}$ be the equivalent dielectric constant which is completely filling the space between the plates. So, capacitance of such capacitor will be

$C^{'} = \frac{K_{e q} ϵ_{0} A}{t_{1} + t_{2}} (∵ d = t_{1} + t_{2})$

Since, the capacitance in both the cases are same,

$∴ C^{'} = C_{e q}$

$\Rightarrow \frac{K_{e q} ϵ_{0} A}{t_{1} + t_{2}} = \frac{K_{1} K_{2} ϵ_{0} A}{K_{1} t_{2} + K_{2} t_{1}}$

$∴ K_{e q} = \frac{K_{1} K_{2} (t_{1} + t_{2})}{K_{1} t_{2} + K_{2} t_{1}}$

Hence, option $(B)$ is correct.

Alternate Solution:
$K_{e q} = \frac{d}{(t_{1} / K_{1}) + (t_{2} / K_{2})} = \frac{K_{1} K_{2} d}{K_{2} t_{1} + K_{1} t_{2}}$

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