Practice portal › Electric Potential and Capacitance › Capacitance and Dielectrics
Asked in GSEB Board July 2021 · Capacitance of a conductor and a capacitor
Given: A=1 m², d=1 mm=1×10⁻³ m, vacuum (air) between the plates.
Idea: C=(ε₀A)/d with ε₀=8.85×10⁻¹² F m⁻¹.
C=(8.85×10⁻¹²×1)/(1×10⁻³).
C=8.85×10⁻⁹ F, about 8.85 nF.
Even a square metre of plate a millimetre apart gives only nanofarads — which is why practical capacitors use thin dielectrics and rolled foils.
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