A 12.6-kV potential difference is applied across parallel conducting plates with an area of 0.631 m². The airbetween the plates will break down and begin to conduct if the electric field exceeeds about 3 x 106 V/m.(a) How close together can you put the plates and have them hold a charge?dminmm(b) What will the capacitance of the plates be at this distance?Cmax =nF(c) How much charge can the plates hold before the air begins to break down?Qmax = |μC

Given potential difference ∆V=12.6 kV Area of plate (A) =0.631m2. Electric field (E) =3 × 106 V/m

Answered: A 12.6-kV potential difference is…

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