A 12.6-kV potential difference is applied across parallel conducting plates with an area of 0.631 m². The air between 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? dmin mm (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

A 12.6-kV potential difference is applied across parallel conducting plates with an area of 0.631 m². The air between 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? dmin mm (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