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GAUSS'S LAW 27.1. V i a l i e : As discussed in Section 27.1, the symmetry of the electric field must match the symmetry of the charge distribution. In particular, the electric field of a cylindrically symmetric charge distribution cannot have a component parallel to the cylinder axis. Also, the electric field of a cylindrically symmetric charge distribution cannot have a component tangent to the circular cross section. The only shape for the electric field that matches the symmetry of the charge distribution with respect to (i) translation parallel to the cylinder axis, (ii) rotation by an angle about the cylinder axis, and (iii) reflections in any plane containing or perpendicular to the cylinder axis is the one shown in the figure. 27.2. Visualize: t +++++++++++++++++ +++++++++++++++++ 1 ! 1 1 ! 1 1 1 1 1 1 1 1 1 1 Figure 27.6 shows the electric field for an infinite plane of charge. For two parallel planes, this is the only shape of the electric field vectors that matches the symmetry of the charge distribution. 27.4. positive charge and inro a closed surface surrounding a net negative charge. Visualize: Solve: line perpendiculur to the plane of the surface. The electric flux our of the closed cube surface is Model: The electric flux "flows" our of a closed surface around a region of space containing a net Please refer to Figure Ex27.4. Let A be the area in m2 of each of the six faces of the cube. The electric flux is defined as = .A = EAcose, where 8 is the angle between the electric field and a (Pa", = (20 N / C + 20 N / C + 10 N /C)AcosO" = (50A) N m' / C Similarly. the electric tlux inro the closed cube surface is = (15 N / C + 15 N / C + 15 N / C)Acos180 = - ( U A ) N m' / C The net electric flux is (50A) N m' / C - (45A) N m' / C = (5A) N m' / C . Since the net electric tlux is positive (i.e.. outward). the closed box contains a positive charge. 27-1 27-2 Chapter 27 27.5. positive charge and into a closed surface surrounding a net negative charge. Visualize: Please refer to Figure Ex27.5. Let A be the area of each of the six faces of the cube. Solve: line perpendicular to the plane of the surface. The electric flux out of the closed cube surface is Qoul = (10 N / C +10 N / C+lON / C + 5 N / C)AcosO" = (35A) N m2 / C Model: The electric flux "flows" out of a closed surface around a region of space containing a net The electric flux is defined as Qe = E . 2 = EAcose, where 8 is the angle between the electric field and a Similarly, the electric flux into the closed cube surface is Qin =(15 N/C+20N/C)Acos(l8O0)=-(35A)Nm2 / C Hence, Qoul + Qin = 0 N m2/C. Since the net electric flux is zero, the closed box contains no charge. 27.6. positive charge and into a closed surface surrounding a net negative charge. ... View Full Document

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