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Standard XII

Physics

Introducing Epsilon

A parallel pl...

Question

A parallel plate condenser is filled with two dielectrics as shown.

Area of each plate is $A$ $m^{2}$ and the separation is $t$ m. The dielectric constants are $k_{1}$ and $k_{2}$ respectively. Its capacitance in farad will be

\frac{ε_{o} A}{t} (k_{1} + k_{2})

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\frac{ε_{o} A}{2 t} (k_{1} + k_{2})

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\frac{2 ε_{o} A}{t} (k_{1} + k_{2})

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\frac{ε_{o} A}{2 t} (k_{1} - k_{2})

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Solution

The correct option is B $\frac{ε_{o} A}{2 t} (k_{1} + k_{2})$
Two dielectrics act as the two capacitors in parallel. So capacitance is given by
$C = C_{1} + C_{2} = \frac{k_{1} ε_{o} A}{2 t} + \frac{k_{2} ε_{o} A}{2 t} = \frac{ε_{o} A}{2 t} (k_{1} + k_{2})$

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Area of each plate is $A$ $m^{2}$ and the separation is $t$ m. The dielectric constants are $k_{1}$ and $k_{2}$ respectively. Its capacitance in farad will be

The correct option is B $\frac{ε_{o} A}{2 t} (k_{1} + k_{2})$
Two dielectrics act as the two capacitors in parallel. So capacitance is given by
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