A parallel plate capacitor having cross-sectional area $A$ and separation $d$ has air in between the plates. Now an insulating slab of the same area but the thickness, $\frac{d}{2}$ , is inserted between the plates as shown in figure having dielectric constant $K = 4 .$ The ratio of new capacitance to its original capacitance will be,

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Aicharya · Accepted Answer

First, recall the formula of capacity of a parallel plate capacitor at free space in terms of cross-section area and separation between plates, $C_{0} = \frac{ϵ_{0} A}{d}$ , when a dielectric slabs of thickness, $t$ and dielectric constant $k$ , inserted between plates then-new capacitance becomes,

$C_{k} = \frac{ϵ_{0} A}{d - t + \frac{t}{k}}$ $\Rightarrow C_{k} = \frac{ϵ_{0} A}{d - \frac{d}{2} + \frac{d}{8}}$ , here $t = \frac{d}{2}$ .

$\Rightarrow C_{k} = \frac{8}{5} \frac{ϵ_{0} A}{d}$ $\Rightarrow C_{k} = \frac{8}{5} C_{0}$ $\Rightarrow \frac{C_{k}}{C_{0}} = \frac{8}{5}$

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A parallel plate capacitor having cross-sectional area A and separation d has air in between the plates. Now…

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