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Idea behind a PPC

Three slabs o...

Question

Three slabs of same dimension having dielectric constant $K_{1}, K_{2}$ and $K_{3}$ completely fills the space between the plates of a capacitor as shown in figure. If $C$ is the initial capacitance of the capacitor, the new capacitance is?

(\frac{K_{1} K_{2} K_{3}}{K_{2} K_{3} + K_{1} K_{3} + K_{1} K_{2}}) 3 C

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

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(\frac{K_{2} K_{3} + K_{1} + K_{3} + K_{1} K_{2}}{K_{1} K_{2} K_{3}}) 3 C

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(\frac{K_{1} K_{2} K_{3}}{K_{1} + K_{2} + K_{3}}) C

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Solution

The correct option is A $(\frac{K_{1} K_{2} K_{3}}{K_{2} K_{3} + K_{1} K_{3} + K_{1} K_{2}}) 3 C$
Let the capacitance of dielectric slabs $K_{1}, K_{2}$ and $K_{3}$ be $C_{1}, C_{2}$ and $C_{3}$ respectively.

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

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

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

From the given figure it is clear that $C_{1}, C_{2}$ and $C_{3}$ are in series.

Now capacitance of the new system is,

$\frac{1}{C_{n e w}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$

$\frac{1}{C_{n e w}} = \frac{d}{3 K_{1} ε_{0} A} + \frac{d}{3 K_{2} ε_{0} A} + \frac{d}{3 K_{3} ε_{0} A}$

$\frac{1}{C_{n e w}} = \frac{1}{3 K_{1} C} + \frac{1}{3 K_{2} C} + \frac{1}{3 K_{3} C}$

$\frac{1}{C_{n e w}} = \frac{1}{3 C} [\frac{1}{K_{1}} + \frac{1}{K_{2}} + \frac{1}{K_{3}}]$

$C_{n e w} = [\frac{K_{1} K_{2} K_{3}}{K_{2} K_{3} + K_{1} K_{3} + K_{1} K_{2}}] 3 C$

Hence, option $(a)$ is the correct answer.

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Idea behind a PPC

Standard XII Physics

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Three slabs of same dimension having dielectric constant K1,K2 and K3 completely fills the space between the plates of a capacitor as shown in figure. If C is the initial capacitance of the capacitor, the new capacitance is?

Three slabs of same dimension having dielectric constant $K_{1}, K_{2}$ and $K_{3}$ completely fills the space between the plates of a capacitor as shown in figure. If $C$ is the initial capacitance of the capacitor, the new capacitance is?