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 capacitance of the capacitor. 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 decreases and the dielectric constant increases. This causes the capacitance of the capacitor to decrease and the electric field between the plates to decrease.
