Gauss's law relates the electric flux @E through a closed surface to the total charge gem] enclosed by the surface: " " 9 (:1 e3 = 36 E - (M = :1; . 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 field from a symmetric charge distribution. The simplest case in which Gauss's law can be used to determine the electric field is that in which the charge is localized at a point, a line, or a plane. When the charge is localized at a point, so that the electric field radiates in three dimensional space, the Gaussian surface is a sphere, and computations can be done in spherical coordinates. Now consider extending all elements of the problem (charge, Gaussian surface, boundary conditions) infinitely along some direction, say along the zaxis. In this case, the point has been extended to a line, namely, the zaxis, and the resulting electric field has cylindrical symmetry. Consequently, the problem reduces to two dimensions, since the field varies only with x and y, or with r and 6 in cylindrical coordinates. A one-dimensional problem may be achieved by extending the problem uniformly in two directions. In this case, the point is extended to a plane, and consequently, it has planar symmetry. Figure ( 2 of 3 Gaussian surface V Part B By symmetry, the electric field must point radially outward from the wire at each point; that is, the field lines lie in planes perpendicular to the wire. In solving for the magnitude of the radial electric field E (r) produced by a line charge with charge density ), one should use a cylindrical Gaussian surface whose axis is the line charge. The length of the cylindrical surface L should cancel out of the expression for E (r) . Apply Gauss's law to this situation to find an expression for E (r). (Figure 2) Express E (r) in terms of some or all of the variables ), r, and any needed constants. View Available Hint(s) IVO AEd ? Er = 2Ter Submit Previous Answers X Incorrect; Try Again; 2 attempts remaining0 =7 ar Oand () =7 EQAxl = enc to qenc = charge enclosed 1 = change ( 9) 9 = 41 length (e ) Eanal = de to E = 1 ( 8 ) in radial outward diretion 2 = unit vector ECO) = 1