A slab of dielectric constant \(K\) has the same crosssectional area as the plates of a parallel plate capacitor and thickness \(\frac{3}{4}\,d\), where \(d\) is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be.(Given \(C _{0}=\) capacitance of capacitor with air as medium between plates.)

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.