JEE Main 2022 Physics question
A slab of dielectric constant K has the same cross-sectional area as the plates of a parallel plate capacitor and thickness 3d/4, where d is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be: (Given = capacitance of capacitor with air as medium between plates.)
Answer: (1)
Solution
+ t/K) with t = 3d/4: + 3d/(4K)) = .