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 $..........\mu F$
A$5$
B$6$
C$4$
D$3$
JEE MAIN 2023, Medium
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B$6$
b $C _{\text {new }}=\frac{\in_0 A }{\frac{\left(\frac{ d }{2}\right)}{1.5}+\frac{\left(\frac{ d }{2}\right)}{1}}$
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