Question 95 of 180

A parallel-plate capacitor with plate area $A$ has separation $d$ between the plates. Two dielectric slabs of dielectric constant ${K}_{1}$ and ${K}_{2}$ of same area $\dfrac{A}{2}$ and thickness $\dfrac{d}{2}$ are inserted in the space between the plates. The capacitance of the capacitor will be given by :

A$\dfrac{{\epsilon }_{0}A}{d}\left(\dfrac{1}{2}+\dfrac{{K}_{1}{K}_{2}}{{K}_{1}+{K}_{2}}\right)$B$\dfrac{{\epsilon }_{0}A}{d}\left(\dfrac{1}{2}+\dfrac{2\left({K}_{1}+{K}_{2}\right)}{{K}_{1}{K}_{2}}\right)$C$\dfrac{{\epsilon }_{0}A}{d}\left(\dfrac{1}{2}+\dfrac{{K}_{1}+{K}_{2}}{{K}_{1}{K}_{2}}\right)$D$\dfrac{{\epsilon }_{0}A}{d}\left(\dfrac{1}{2}+\dfrac{{K}_{1}{K}_{2}}{2\left({K}_{1}+{K}_{2}\right)}\right)$

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Answer: (A) $\dfrac{{\epsilon }_{0}A}{d}\left(\dfrac{1}{2}+\dfrac{{K}_{1}{K}_{2}}{{K}_{1}+{K}_{2}}\right)$

${C}_{eq}={C}_{1}+\dfrac{{C}_{2}{C}_{3}}{{C}_{2}+{C}_{3}}$
${C}_{eq}=\dfrac{A{\epsilon }_{0}}{2d}+\dfrac{\dfrac{{K}_{1}A{\epsilon }_{0}}{d}\times \dfrac{{K}_{2}A{\epsilon }_{0}}{d}}{\dfrac{{K}_{1}A{\epsilon }_{0}}{d}+\dfrac{{K}_{2}A{\epsilon }_{0}}{d}}$
${C}_{eq}=\dfrac{A{\epsilon }_{0}}{2d}+\dfrac{A{\epsilon }_{0}}{d}\left(\dfrac{{K}_{1}{K}_{2}}{{K}_{1}+{K}_{2}}\right)=\dfrac{A{\epsilon }_{0}}{d}\left(\dfrac{1}{2}+\dfrac{{K}_{1}{K}_{2}}{{K}_{1}+{K}_{2}}\right)$