A charge of 6.2 `microC resides on a wire 38 meters long.
What is the electric field at a point .21 meters from the wire?
The total flux is 4`pi kq, where q is the total charge 6.2 `microC. This flux is 4`pi (9 x 10^9 N m^2/C^2 ( 6.2 `microC)) = 701200 N m^2/C.
The total surface area of the cylinder is 38 meters (2`pi ( .17 m))) = 40.58 m^2.
The magnitude of the electric field is flux / area = ( 701200 N m^2 / C) / ( 40.58 m^2 ) ) = 17270 N / C.
At a distance of .21 meters from the wire, the flux would be evenly spread over a cylinder 38 m long and with radius .21 meters.
- electric field = flux / area = ( 701200 N m^2 / C) / ( 50.13 m^2) = 13980 N / C.
A total charge Q will produce flux 4 `pi k Q.
If the charge is evenly distributed on a wire of length L, then a cylinder of radius r and length L will have total area 2 `pi r L on its curved surface.
If the cylinder is uniformly penetrated by all the flux produced by the charge, the flux density on this curved surface will be the electric field strength:
This can be written as
where the Greek letter `lambda is the charge / unit length along the wire.
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