28 1 Visualize As discussed in Section 28 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 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 respect with 28 2 Visualize The object has spherical symmetry so the electric field is radial 28 3 Visualize Figure 28 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 28 4 Model The electric flux flows out of a closed surface around a region of space containing a net positive charge and into a closed surface surrounding a net negative charge Visualize Please refer to Figure EX28 4 Let A be the area in m2 of each of the six faces of the cube Solve The electric flux is defined as a line perpendicular to the plane of the surface The electric flux out of the closed cube surface is where is the angle between the electric field and cid 71 cid 71 E A EA cos e out 20 N C 20 N C 10 N C cos0 A 50 A N m C 2 Similarly the electric flux into the closed cube surface is in 15 N C 15 N C 15 N C cos180 A 45 A N m C 2 The net electric flux is i e outward the closed box contains a positive charge N m C 45 N m C 50 A A 2 2 5 A 2 N m C Since the net electric flux is positive 28 5 Model The electric flux flows out of a closed surface around a region of space containing a net positive charge and into a closed surface surrounding a net negative charge Visualize Please refer to Figure EX28 5 Let A be the area of each of the six faces of the cube cid 71 cid 71 E A EA cos Solve The electric flux is defined as a line perpendicular to the plane of the surface The electric flux out of the closed cube surface is where is the angle between the electric field and e out 10 N C 10 N C 20 N C 5 N C cos0 A 45 A N m C 2 Similarly the electric flux into the closed cube surface is in 15 N C 20 N C cos 180 A 35 A N m C 2 Hence out in 10 N m2 C Since the net electric flux is positive i e outward the closed box contains a positive charge 28 6 Model The electric flux flows out of a closed surface around a region of space containing a net positive charge and into a closed surface surrounding a net negative charge Visualize Please refer to Figure EX28 6 Let A be the area of each of the six faces of the cube cid 71 cid 71 E A EA cos Solve The electric flux is defined as a line perpendicular to the plane of the surface The electric flux out of the closed cube surface is where is the angle between the electric field and e out 20 N C 10 N C 10 N C cos0 A 40 A N m C 2 Similarly the electric flux into the closed cube surface is in 20 N C 15 N C cos180 A 35 A N m C 2 Because the cube contains negative charge out in must be negative This means out in unknown 0 N m2 C Therefore 40 A N m C 2 35 A N m C 2 unknown 0 N m C 2 unknown 5 A N m C 2 That is the unknown vector points into the front face of the cube and its field strength is greater than 5 N C 28 7 Model The electric flux flows out of a closed surface around a region of space containing a net positive charge and into a closed surface surrounding a net negative charge Visualize Please refer to Figure EX28 7 Let A be the area of each of the six faces of the cube cid 71 cid 71 E A EA cos Solve The electric flux is defined as a line perpendicular to the plane of the surface The electric flux out of the closed cube surface is where is the angle between the electric field and e out 20 N C 10 N C cos0 A 30 A N m C 2 Similarly the electric flux into the closed cube surface is Because the cube contains positive charge out in must be positive This means out in unknown 0 N m2 C Therefore 20 N C 15 N C 10 N C cos180 N m C in 45 A A 2 30 A N m C 2 45 A N m C 2 unknown 0 N m C 2 unknown 15 A N m C 2 That is the unknown vector points out of the front face of the cube and its field strength is greater than 15 N C 28 8 Model The electric flux flows out of a closed surface around a region of space containing a net positive charge and into a closed surface surrounding a net negative charge Visualize Please refer to Figure EX28 8 Let A be the area of each of the six faces of the cube cid 71 cid 71 E A EA cos Solve The electric flux is defined as a line perpendicular to the plane of the surface The electric flux out of the closed cube surface is where is the angle between the electric field and e out 15 N C 15 N C 10 N C cos0 A 40 A N m C 2 Similarly the electric flux into the closed cube surface is in 20 N C 15 N C cos180 A 35 A N m C 2 Because the cube contains no charge the total flux is zero This means in out unknown 0 N m2 C unknown 5 A N m C 2 The unknown electric field vector on the front face points in and the electric field strength is 5 N C 28 9 Model The electric field is uniform over the entire surface Visualize Please refer to Figure EX28 9 The electric field vectors make an angle of 30 with the planar surface Because the normal n to the planar surface is at an angle of 90 with the surface the angle between n cid 71 and E Solve The electric flux is 60 is cid 71 cid 71 E A EA cos …
View Full Document
Unlocking...