A parallel plate capacitor having crosssectional area $A$ and separation $d$ has air in between the plates. Now an insulating slab of same area but thickness $d/2$ 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,
NEET 2020, Medium
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$C_{0}=\frac{\varepsilon_{0} A}{d}$
After inserting dielectric
$C=\frac{\varepsilon_{0} A}{(d-t)+\frac{t}{k}}$
$=\frac{\varepsilon_{0} A}{\frac{d}{2}+\frac{d}{8}}$
$=\frac{8 \varepsilon_{0} A}{5 d}$
$=\frac{8}{5} C_{0}$
So, $\frac{C}{C_{0}}=\frac{8}{5}$
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