Practice question
Question
A parallel plate capacitor with \( C = 40 \, \text{pF} \) in air has a dielectric (\( K = 5 \))
inserted fully between plates. What is the new capacitance?
Explanation
**Effect of dielectric** is to reduce effective field due to polarization, bound surface charges opposite to free charges, net field E = E₀ - E_p = E₀/K. Capacitance C = Q/V = K ε₀ A/d for full filling, partial filling C = ε₀ A/(d - t + t/K) for slab thickness t. C' = K C = 5 × 40 = 200 pF . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C V² and common potential V = Q_total/C_total, result 200 pF follows, reflecting potential-capacitance relations.
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.