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  • A parallel-plate capacitor  with plate area A has seperation d between the plates. Two dielectric slabs of dielectric constant K1 and K2 of same A/2 and thickness d/2 are inserted in the space between the plates. The capacitance

A parallel-plate capacitor  with plate area A has seperation d between the plates. Two dielectric slabs of dielectric constant K1 and K2 of same A/2 and thickness d/2 are inserted in the space between the plates. The capacitance of the capacitor will be given by :

Option: 1 \frac{\varepsilon _{0}A}{d} \left ( \frac{1}{2}+\frac{K_{1}K_{2}}{K_{1}+K_{2}} \right )
Option: 2 \frac{\varepsilon _{0}A}{d} \left ( \frac{1}{2}+\frac{2\left ( K_{1}+K_{2} \right )}{K_{1}K_{2}}\right )
Option: 3 \frac{\varepsilon _{0}A}{d} \left ( \frac{1}{2}+\frac{K_{1}+K_{2}}{K_{1}K_{2}} \right )
Option: 4 \frac{\varepsilon _{0}A}{d} \left ( \frac{1}{2}+\frac{K_{1}K_{2}}{2\left ( K_{1}+K_{2} \right )} \right )

Answers (1)

best_answer



C_{eq}= \frac{c_{1}c_{2}}{c_{1}+c_{2}}+C
       = \frac{\left ( \frac{A}{2} \right )\varepsilon _{0}}{\frac{\frac{d}{2}}{k_{1}}+\frac{\frac{d}{2}}{k_{2}}}+\frac{A}{2}\frac{\varepsilon _{0}}{d}
C_{eq}= \frac{A\varepsilon _{0}\left ( k_{1}k_{2} \right )}{\left ( k_{2}+k_{1} \right )d}+\frac{A\varepsilon _{0}}{2d}
\Rightarrow C_{eq}=\frac{A\varepsilon _{0}}{d} \left [ \frac{1}{2}+\frac{k_{1}k_{2}}{k_{1}+k_{2}}\right ]
The correct option is (1)

Posted by

vishal kumar

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