A capacitor has capacitance 5 µ F 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 (d)/(2) is

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A capacitor has capacitance 5 µ F when its parallel plates are separated by air medium of thickness d . A slab of

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

A capacitor has capacitance 5 µ F when its parallel plates are separated by air medium of thickness d . A slab of

After insertion of dielectric, the equivalent capacitance of the new capacitor is

${C}_{eq} = \dfrac{{ϵ}_{0}A}{\dfrac{\dfrac{d}{2}}{1 . 5} + \dfrac{\dfrac{d}{2}}{1}}$

$= \dfrac{{ε}_{0}A}{\left(\dfrac{d}{3} + \dfrac{d}{2}\right)} = \dfrac{6{ε}_{0}A}{5d}$

$= \dfrac{6}{5} \times 5\mu F = 6\mu F$ .

A capacitor has capacitance 5 µ F when its parallel plates are...

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