A flat square sheet of thin aluminum foil, 25cm on a side carries a uniformly distributed 35nC chargeWhat, approximately is the electric field (a) 1.0 cm above the sheet and (b) 20m above the sheet?Should I calculate the electric field (σA)/ε0 which is: [(3510^-9C)(.25m)^2] / 8.8510^-12 2.510^2If that is correct how do I find out what the field is at each distance.Do I use the E(kQ)/(r^2) formula? When using this it seems to generate a larger field at 1cm (3.1510^6) than the field at the origin 2.1510^2 and 20m gives 7.9 10^-1Any help is appreciated.
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(a) The 1 cm is irrelevant per se other than telling you that this is in the near field (1cm 25cm)That means the electric field is nearly the same as for an infinite plane with uniform surface density - E (1/2)(Charge/Area)(1/epsilon0) in MKS (b) At 20 m it's the same as the electric field due to a point charge of magnitude 35 nC (q) at a distance of 20 meters (r) - E q/r^2 (1/4pi epsilon0) in MKS
