Practice question
Question
A parallel plate capacitor with capacitance \( 50 \, \text{pF} \) has a dielectric (\( K = 2 \),
thickness \( d/3 \)) inserted. What is the new capacitance? (Original separation \( d \)).
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. Potential difference: V = E₀ ( (2d/3) ) + (E₀/K) ( (d/3) ) = E₀ d ( (2/3) + (1/3 × 2) ) = E₀ d ( (2/3) + (1/6) ) = E₀ d × (5/6) . C = (Q/V) = (Q/(5/6) V₀) = (6/5) × 50 = 60 pF . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C =
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