Previous Year NEET Question- Displacement Current

1.

Given:

C = 8pF d = d K = 5

We know that

C = εA/d

So,

C1 = εA/d C2 = ε(A/2)/d/2 = 5*εA/d = 5C1

Therefore, the capacitance of the capacitor is now 5C₁ = 40pF.

2.

The correct answer is B.

When a dielectric slab is inserted between the plates of a capacitor, the electric field between the plates decreases. This is because the dielectric slab reduces the electric field, not the capacitance. The capacitance of a capacitor is inversely proportional to the distance between the plates and directly proportional to the dielectric constant of the material between the plates. When a dielectric slab is inserted between the plates of a capacitor, the distance between the plates remains the same and the dielectric constant increases. This causes the capacitance of the capacitor to increase and the electric field between the plates to decrease.