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

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.