27 1 Model The electric field is that of the two charges placed on the y axis Visualize Please refer to Figure EX27 1 We denote the upper charge by q1 and the lower charge by q2 Because both the charges are positive their electric fields at P are directed away from the charges Solve The electric field from q1 is cid 71 E 1 1 4 0 q 1 2 r 1 below x axis 9 2 9 0 10 N m C 3 0 10 C 0 050 m 0 050 m 9 2 2 2 cos i sin j Because tan 5 cm 5 cm 1 the angle 45 Hence Similarly the electric field from q2 is cid 71 E 1 5400 N C 1 2 i 1 2 j cid 71 E 2 1 4 0 q 2 2 r 2 above x axis 5400 N C 1 2 i 1 2 j cid 71 E net at P cid 71 cid 71 E E 1 2 2 5400 N C i 3 7 6 10 N C i 1 2 Thus the strength of the electric field is 7 6 103 N C and its direction is horizontal Assess Because the charges are located symmetrically on either side of the x axis and are of equal value the y components of their fields will cancel when added 27 2 Model The electric field at the point is found by superposition of the fields due to the two charges located on the y axis Visualize Please refer to Figure EX27 2 The electric field due to the positive charge q1 at the point is away from q1 On the other hand the electric field due to the negative charge q2 at the point is toward q2 These two electric fields are then added vertically to obtain the net electric field at the point Solve The electric field from q1 is cid 71 E 1 1 4 0 q 1 2 r 1 below axis x 2 9 9 0 10 N m C 3 0 10 C 0 050 m 0 050 m 9 2 2 2 cos i sin j Because tan 5 cm 5 cm 45 So Similarly the electric field from q2 is cid 71 E 1 5400 N C i 1 2 1 2 j cid 71 E 2 1 4 0 q 2 2 r 2 cid 71 E net cid 71 cid 71 E E 1 2 below x axis 5400 N C i 1 2 1 2 j 2 5400 N C j j 7637 N C 3 7 6 10 N C 1 2 Thus the strength of the electric field is 7 6 103 N C and its direction is vertically downward Assess A quick visualization of the components of the two electric fields shows that the horizontal components cancel 27 3 Model The electric field is that due to superposition of the fields of the two 3 0 nC charges located on the y axis Visualize Please refer to Figure EX27 3 We denote the top 3 0 nC charge by q1 and the bottom 3 0 nC charge of both the positive charges are directed away from their respective by q2 The electric fields and cid 71 E cid 71 E 1 2 charges With vector addition they yield the net electric field Solve The electric fields from q1 and q2 are cid 71 netE at the point P indicated by the dot cid 71 E 1 q 1 1 2 4 r 0 1 along x axis 9 0 10 N m C 3 0 10 C 9 9 2 2 0 05 m 2 i i 10 800 N C cid 71 E 2 1 4 0 q 2 2 r 2 above x axis Because tan 10 cm 5 cm tan 1 2 63 43 So cid 71 E 2 2 9 9 0 10 N m C 3 0 10 C 0 050 m 0 10 m 9 2 2 2 i cos63 43 j sin 63 43 i 966 j 1127 N C The net electric field is thus To find the angle this net vector makes with the x axis we calculate cid 71 E net at P cid 71 E 1 cid 71 E 2 i 11 766 1127 N C j tan 1127 N C 11 766 N C 5 5 Thus the strength of the electric field at P is E net 11 766 N C 1127 N C 2 2 11 820 N C 1 18 103 N C cid 71 netE and makes an angle of 5 5 above the x axis Assess Because of the inverse square dependence on distance no special symmetry relative to the charges we expected the net field to be at an angle relative to the x axis Additionally because the point P has E 1 E 2 27 4 Model The electric field at the point is the superposition of the fields due to the two charges located on the y axis Visualize Please refer to Figure EX27 4 We denote the positive charge by q1 and the negative charge by q2 is toward the negative The electric field of the positive charge q1 is directed away from q1 but the field cid 71 2E cid 71 1E vectorially to find the strength and the direction of the net electric field charge q2 We will add vector Solve The electric fields from q1 and q2 are and cid 71 1E cid 71 2E cid 71 E 1 q 1 1 2 r 4 0 1 q away from 1 along axis x 9 0 10 N m C 3 0 10 C 9 9 2 2 0 050 m 2 i i 10 800 N C cid 71 E 2 1 4 0 q 2 2 r 2 below axis x From the geometry of the figure tan 63 43 10 cm 5 cm cos 63 43 i sin 63 43 i 966 1127 j j N C cid 71 E 2 9 2 9 0 10 N m C 3 0 10 C 0 050 m 0 10 m 9 2 2 2 cid 71 cid 71 E E 1 cid 71 E net 2 i 9834 1127 j N C E net 9834 N C 2 1127 N C 2 9 9 103 N C To find the angle this net vector makes with the horizontal we calculate tan E net E net y x 1127 N C 9834 N C 6 5 Thus the strength of the net electric field at the point is 9 9 103 N C and x axis cid 71 netE makes an angle of 6 5 below the 27 5 Model The distances to the observation points are large compared to the size of the dipole so model the field as that of a dipole moment Visualize The dipole consists of charges q along the y axis The electric field in a points down The field in b points up Solve a The dipole moment is cid 71 p qs …
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