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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. 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)

{{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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