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 \(..........\mu F\)

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(B) \(6\)

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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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