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. Potential difference: V = E₀ ( (7d/8) ) + (E₀/K) ( (d/8) ) = E₀ d ( (7/8) + (1/8 × 3) ) . V = E₀ d ( (7/8) + (1/24) ) = E₀ d × (22/24) = E₀ d × (11/12) . C = (Q/V) = (Q/(11/12) V₀) = (12/11) × 300 ≈ 327.27 pF