This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: ASSIGNMENT 4 21.5.I DENTIFY : Apply 1 2 2 k q q F r = and solve for r . S ET U P : 650 N F = . E XECUTE : 9 2 2 2 1 2 3 (8.99 10 N m /C )(1.0 C) 3.7 10 m 3.7 km 650 N k q q r F = = = = E VALUATE : Charged objects typically have net charges much less than 1 C. 21.44. I DENTIFY : For a point charge, 2 q E k r = . For the net electric field to be zero, 1 E r and 2 E r must have equal magnitudes and opposite directions. S ET U P : Let 1 0.500 nC q = + and 2 8.00 nC. q = + E r is toward a negative charge and away from a positive charge. E XECUTE : The two charges and the directions of their electric fields in three regions are shown in Figure 21.44. Only in region II are the two electric fields in opposite directions. Consider a point a distance x from 1 q so a distance 1.20 m x- from 2 q . 1 2 E E = gives 2 2 0.500 nC 8.00 nC (1.20 ) k k x x =- . 2 2 16 (1.20 m ) x x =- . 4 (1.20 m ) x x = - and 0.24 m x = is the positive solution. The electric field is zero at a point between the two charges, 0.24 m from the 0.500 nC charge and 0.96 m from the 8.00 nC charge. E VALUATE : There is only one point along the line connecting the two charges where the net electric field is zero. This point is closer to the charge that has the smaller magnitude. Figure 21.44 21.49. I DENTIFY : The electric field of a positive charge is directed radially outward from the charge and has magnitude 2 1 ....
View Full Document
This note was uploaded on 03/11/2010 for the course PHYSICS 1523 taught by Professor Dfs during the Fall '10 term at Dallas.
- Fall '10