A flat square sheet of thin aluminum foil, 25cm on a side carries a uniformly distributed 35nC charge. What, 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: [(35*10^-9C)(.25m)^2] / 8.85*10^-12 = 2.5*10^2
If 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 (3.15*10^6) than the field at the origin 2.15*10^2 and 20m gives 7.9 *10^-1.
Any help is appreciated...
2006-09-04 02:43:30 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
(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
2006-09-04 08:34:17 · answer #1 · answered by shimrod 4 · 0⤊ 0⤋
I think you are correct, and don't worry about the origin. It is a boundary condition--the field is not defined there.
2006-09-04 09:54:00 · answer #2 · answered by Speedy 3 · 0⤊ 0⤋