Practice question
Question
A parallel plate capacitor with \( C = 100 \, \text{pF} \) in air has a dielectric (\( K = 10 \))
inserted fully between plates. What is the new capacitance?
Explanation
**Capacitance depends on geometry** not charge, C = Q/V constant for given arrangement. Parallel plate C ∝ A/d, so larger area and smaller separation increase capacitance, principle used to increase storage by using large area foils with thin dielectric. C' = K C = 10 × 100 = 1000 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 1000 pF follows, reflecting potential-capacitance relations.
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