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Byju's Answer

Standard XII

Physics

Towards Biot-Savart Law

A parallel pl...

Question

A parallel plate capacitor has the space between its plates filled by two slabs of thickness $d / 2$ each with dielectric constants $k_{1}$ and $k_{2}$ . Given that $d$ is the plate separation of the capacitor, the effective capacitance is

\frac{ϵ_{0} A (k_{1} + k_{2})}{2 d}

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\frac{2 ϵ_{0} A}{d} \frac{k_{1} k_{2}}{k_{1} + k_{2}}

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\frac{ϵ_{0} A}{d} \frac{k_{1} k_{2}}{k_{1} - k_{2}}

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\frac{ϵ_{0} A (k_{1} - k_{2})}{d}

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Solution

The correct option is B $\frac{2 ϵ_{0} A}{d} \frac{k_{1} k_{2}}{k_{1} + k_{2}}$
The parts behave like 2 capacitors connected in series

This implies,
$C_{1} = \frac{k_{1} ϵ_{0} A}{d / 2} = \frac{2 k_{1} ϵ_{0} A}{d}$ and

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

Now the effective capacitance for series combination,

$\frac{1}{C_{e q}} = \frac{1}{C_{1}} + \frac{1}{C_{2}}$

$\Rightarrow C_{e q} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = \frac{2 ϵ_{0} A}{d} \frac{k_{1} k_{2}}{(k_{1} + k_{2})}$

Hence, option $(b)$ is the correct choice.

Key concept- Two or more dielectrics placed parallel to the capacitor plates can be assumed to be in series.

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