Practice portal › Electric Potential and Capacitance › Capacitor Circuits and Energy
Asked in GUJCET 2019 · Capacitor networks
Given: four plates 1–4 of area A, each gap d; plates 1 and 4 are joined to A, plates 2 and 3 to B.
Idea: each gap between facing plates at different potentials is a capacitor C = (ε₀A)/d.
Gap 1–2 lies between A and B: a capacitor C.
Gap 2–3 has both plates at B: no potential difference, so no capacitor.
Gap 3–4 lies between B and A: another C.
The two working gaps are in parallel: C_AB = 2C = (2ε₀A)/d.
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