Practice portal › Electric Potential and Capacitance › Capacitance and Dielectrics
Asked in GSEB Board August 2020 · Capacitance of a conductor and a capacitor
Given: C = 8.85 μF = 8.85×10⁻⁶ F, d = 1 mm = 10⁻³ m, ε₀ = 8.85×10⁻¹² C² N⁻¹ m⁻².
Idea: C = (ε₀A)/d, so A = (Cd)/(ε₀).
A = (8.85×10⁻⁶×10⁻³)/(8.85×10⁻¹²) = 1×10³ m².
That is a plate about 32 m on a side: a microfarad is a very large capacitance for an air-filled parallel plate capacitor.
So A = 1×10³ m².
Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.
Practise this with a timer