Q4. Assertion: with the battery still connected, sliding a dielectric into the gap of a capacitor increases the charge on its plates. Reason: the capacitance increases while the battery holds the potential difference fixed. Are both true, and does the reason explain the assertion?
Both are true, and the reason explains the assertion: Q = CV, and C rises to KC0 while V stays fixed (NCERT Eqs. (2.38) and (2.54), pages 67 and 70).
Q5. A metal slab of thickness d/2, with the same area as the plates, is placed between the plates of a parallel plate capacitor of separation d, without electrically joining the plates; neglect fringing. What is the new capacitance in terms of C0?
The field inside the metal is zero, so only the air gaps, d/2 in all, carry a potential difference: V = E0d/2. So C = 2C0. This is the slab formula with K very large (reasoned from NCERT Example 2.8 and the zero field inside a conductor, page 62).
Q6. A slab of dielectric constant 4, half as thick as the gap, is placed between the plates of a parallel plate capacitor. Find the new capacitance in terms of C0.
C = ε0A/(d − t + t/K) with t = d/2: d − t + t/K = d/2 + d/8 = 5d/8, so C = (8/5)C0 = 1.6C0 (reasoned from NCERT Example 2.8, page 71).