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  • A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness d_{1} and dielectric constant k_{1} and the other has thickness d_{2} and dielectric constant k_{2} as shown in the figure.

A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness d_{1} and dielectric constant k_{1} and the other has thickness d_{2} and dielectric constant k_{2} as shown in the figure. This arrangement can be thought of as a dielectric slab of thickness d=d_{1}+d_{2} and effective dielectric constant k. The k is

a) \frac{2K_{1}K_{2}}{\left ( K_{1}+K_{2} \right )}

b) \frac{K_{1}d_{1}+K_{2}d_{2}}{\left ( d_{1}+d_{2} \right )}

c) \frac{\left ( K_{1}d_{1}+K_{2}d_{2} \right )}{\left ( K_{1}+K_{2} \right )}

d)\frac{K_{1}K_{2}\left ( d_{1}+d_{2} \right )}{\left ( K_{1}d_{1}+K_{2}d_{2} \right )}

Answers (1)

The answer is the option (d)

Explanation:-

When two dielectrics are in series, we treat them as two separate capacitors in series. The effective capacitance of two capacitors in series is given by the reciprocal of the sum of the reciprocals of their individual capacitances.

{{C}_{eq}}=\dfrac{{{k}_{1}}{{k}_{2}}{{\varepsilon }_{0}}A}{{{k}_{1}}{{d}_{2}}+{{k}_{2}}{{d}_{1}}}.............(1)

C=\dfrac{k{{\varepsilon }_{0}}A}{{{d}_{1}}+{{d}_{2}}}.........(2)

From (1) and (2)

\dfrac{{{k}_{1}}{{k}_{2}}{{\varepsilon }_{0}}A}{{{k}_{1}}{{d}_{2}}+{{k}_{2}}{{d}_{1}}}=\dfrac{k{{\varepsilon }_{0}}A}{{{d}_{1}}+{{d}_{2}}}

K=\frac{K_{1}K_{2}\left ( d_{1}+d_{2} \right )}{\left ( K_{1}d_{1}+K_{2}d_{2} \right )}

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infoexpert23

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