Practice question

**Potential difference increase** statement with constant charge is incorrect; V decreases when K>1 inserted. If slab inserted while maintaining constant charge, E reduces to E₀/K, V = E d reduces. If battery connected maintaining constant V, E stays V/d, D = ε E increases, Q increases K times. With a constant charge Q , capacitance C = (ε₀ A/d) without dielectric, and C' = (K ε₀ A/d) with dielectric ( K > 1 ). When the dielectric is fully inserted, C increases, reducing V = (Q/C) . If the slab is partially removed, the effective capacitance decreases (as less dielectric area contributes K ), approaching