Gauss\'s law relates the electric flux phi_E through a closed surface to the tot
ID: 1655341 • Letter: G
Question
Gauss's law relates the electric flux phi_E through a closed surface to the total charge q_ enclosed by the surface. phi_E = contourintegral E vector middot d A vector = You can use Gauss's law to determine the charge enclosed inside a closed surface on which the electric field is known. However, Gauss's law is most frequently used to determine the electric filed from symmetric Now consider the case that the charge has been extended along the z axis. This is generally called a line charge. The usual variable for a charge density (charge per unit length)is lambda, and it has units (in the SI system) of coulombs per meter. By symmetry, the electric field must point radially outward from the wire at each point, that is, the field lines lie in planes perpendicular the . In solving for the magnitude of the radial electric field E(r) produced by a line charge with charge density lambda, one should use cylindrical Gaussian surface whose axis is the line charge. The length of the cylindrical surface L should cancel out of the expression E(r). Apply Gauss's law to this situation to find an expression for E(r). (Figure 1) Express E(r) in terms of some or all of the variables lambda, r, and any needed constants.Explanation / Answer
from Gauss law,
net flux = Qin / e0
E (2 x pi x r x L) = lambda L / e0
E = lambda / (2 pi e0 r)
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