Question 83 of 180

A capacitor has capacitance $5µF$ when its parallel plates are separated by air medium of thickness $d$. A slab of material of dielectric constant $1.5$ having area equal to that of plates but thickness $\dfrac{d}{2}$ is inserted between the plates. Capacitance of the capacitor in the presence of slab will be ______ $µF$.

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Answer: 6

Capacitance of the parallel plate capacitor when air is present $C=\dfrac{{ϵ}_{0}A}{d}=5\mu F$

After insertion of dielectric, the equivalent capacitance of the new capacitor is
${C}_{eq}=\dfrac{{ϵ}_{0}A}{\dfrac{\dfrac{d}{2}}{1.5}+\dfrac{\dfrac{d}{2}}{1}}$
$=\dfrac{{\epsilon }_{0}A}{\left(\dfrac{d}{3}+\dfrac{d}{2}\right)}=\dfrac{6{\epsilon }_{0}A}{5d}$
$=\dfrac{6}{5}\times 5\mu F=6\mu F$.