(a) Find the magnitude and direction of the electric field at radial distance r
ID: 2271391 • Letter: #
Question
(a) Find the magnitude and direction of the electric field at radial distance r = 4.9 cm from the common central axis. (Take radially outward to be positive.)
(b) Find the magnitude and direction of the electric field at r = 8.2 cm, using the same sign convention.
To get the points, use THESE numbers in the calculations, provide thourough explanations of the logic behind how you set up the problem, and dont copy and paste crap from yahoo answers. I will ignore answers that don't come with an explanation.
Two long, charged, thin-walled, concentric cylinders have radii of 3.0 and 6.0 cm. The charge per unit length is 6.50 Times 10-6 C/m on the inner shell and -6.50 times 10-6 C/m on the outer shell.Explanation / Answer
Gauss's Law says that the flux through a closed surface (which is the electric field integrated over the area of the surface) times the permittivity constant is equal to the charge enclosed.
Set up a cylindrical Gaussian surface 1m long with a radius equal to the two radii specified in your problem. You can see immediately that the answer to part(b) is zero, because a cylinder at r = 8.2 cm encloses both the inner and outer shell, and since their charge is equal but opposite the net charge enclosed is zero.
Part (a) however needs a calculation. Because Chegg is so lame that they do not allow insertion of symbols, superscripts, or subscripts when answering, I will use 'e' to represent epsilon sub zero, the permittivity constant.
Since the gaussian surface at 4.9 cm is symmetric, our integral of E over the area is simply E times the area. We can disregard the outside cylinder because it is not enclosed by the surface, and only worry about the smaller r=3.0 cm surface.
We know eEA = q (Gauss's Law) so E = q/eA. The area of a cylinder 1 m long with a radius of 4.9 cm is
(1 m)(0.049 m)(2pi) = 0.3079 m^2
The cylinder encloses 6.5 * 10^-6 C of charge. So
E = (6.5 * 10^-6 C)/(8.854 * 10^-12 C/Vm)*(0.3079 m^2)
= 2.384 * 10^6 V/m
and the direction will be outwards since the enclosed charge is positive.
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