A capacitor has capacitance $5 \mu F$ when it's 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 $\frac{ d }{2}$ is inserted between the plates. Capacitance of the capacitor in the presence of slab will be $..........\mu F$
JEE MAIN 2023, Medium
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$C _{\text {new }}=\frac{\in_0 A }{\frac{\left(\frac{ d }{2}\right)}{1.5}+\frac{\left(\frac{ d }{2}\right)}{1}}$
$=\frac{\in_0 A}{\left(\frac{d}{3}+\frac{d}{2}\right)}=\frac{6 \in_0 A}{5 d}$
$=\frac{6}{5} \times 5 \mu F =6 \mu F$
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