Q.The dielectric constant of the medium between the two plates of a parallel plate capacitor is 2. Its capacitance is 3μF. Now, the separation between the plates is made doubled and in that vacant space a plate of dielectric constant K = 4 is inserted. Now, the capacitance of the capacitor will be

First find ε₀A/d from the original capacitor, then treat the new capacitor as two dielectric slabs (K=2 and K=4, each thickness d) in series across separation 2d. The result is 2 μF. Option (b).

Step 1 — original: C = Kε₀A/d = 2(ε₀A/d) = 3 μF, so ε₀A/d = 1·5 μF.

Step 2 — new geometry: separation doubled to 2d. The original K=2 dielectric occupies thickness d; the newly created vacant space of thickness d is filled with K=4. These act as two capacitors in series.

Step 3 — series of two partially-filled slabs: …