Practice question
Question
A parallel plate capacitor with \( C = 50 \, \text{pF} \) in air has a dielectric (\( K = 4 \))
inserted fully between plates. What is the new capacitance?
Explanation
**Parallel plate capacitor** capacitance C = ε₀ A/d, ε₀=8.85×10⁻¹² F/m, A plate area (m²), d separation (m), for air, with dielectric C = K ε₀ A/d. For A=0.08 m², d=0.4 mm=4×10⁻⁴ m, C=8.85×10⁻¹²×0.08/4×10⁻⁴=1.77×10⁻⁹ F=1.77 nF, illustrating small capacitance for cm separation. C' = K C = 4 × 50 = 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.
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