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Standard XII

Physics

Capacitance of Parallel Plate Capacitor

A system of t...

Question

A system of two parallel plates, each of area A, are separated by distances $d_{1}$ and $d_{2}$ . The space between them is filled with dielectrics of permittivities $ϵ_{1}$ and $ϵ_{2}$ . The permittivity of free space is $ϵ_{0}$ . The equivalent capacitance of the system is :

\frac{ϵ_{1} ϵ_{2} A}{ϵ_{2} d_{1} + ϵ_{1} d_{2}}

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\frac{ϵ_{1} ϵ_{2} ϵ_{0} A}{ϵ_{1} d_{1} + ϵ_{2} d_{2}}

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\frac{ϵ_{0} A}{ϵ_{1} d_{1} + ϵ_{2} d_{2}}

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\frac{ϵ_{0} A}{ϵ_{1} d_{2} + ϵ_{2} d_{1}}

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Solution

The correct option is A $\frac{ϵ_{1} ϵ_{2} A}{ϵ_{2} d_{1} + ϵ_{1} d_{2}}$
There are two capacitors $C_{1}$ and $C_{2}$ and they are in series.
Here, $C_{1} = \frac{A ϵ_{1}}{d_{1}}$ and $C_{2} = \frac{A ϵ_{2}}{d_{2}}$
The equivalent capacitance is $C_{e q} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = \frac{A ϵ_{1} ϵ_{2}}{ϵ_{1} d_{2} + ϵ_{2} d_{1}}$

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A system of two parallel plates, each of area A, are separated by distances d1 and d2. The space between them is filled with dielectrics of permittivities ϵ1 and ϵ2. The permittivity of free space is ϵ0. The equivalent capacitance of the system is :

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